From daf25d5e643563a8d40685314c326ac073c0405d Mon Sep 17 00:00:00 2001
From: Yudong Jin <krahets@163.com>
Date: Fri, 25 Nov 2022 20:12:20 +0800
Subject: [PATCH] Add python codes and for the chapter of computational
 complexity. Update Java codes. Update Contributors.

---
 README.md                                     |   2 +-
 .../leetcode_two_sum.go                       |   3 +
 .../leetcode_two_sum.java                     |  11 +-
 .../space_complexity_types.java               |   4 +-
 .../time_complexity_types.java                |  21 +-
 .../chapter_array_and_linkedlist/array.py     |   1 +
 .../linked_list.py                            |   5 +-
 .../chapter_array_and_linkedlist/list.py      |   5 +-
 .../chapter_array_and_linkedlist/my_list.py   |   5 +-
 .../leetcode_two_sum.py                       |  31 +++
 .../space_complexity_types.py                 |  68 +++++++
 .../time_complexity_types.py                  | 130 +++++++++++++
 .../worst_best_time_complexity.py             |  24 +++
 .../space_complexity.md                       |  92 +++++++--
 .../space_time_tradeoff.md                    |  67 ++++---
 .../time_complexity.md                        | 182 +++++++++++++++---
 docs/chapter_preface/index.md                 |   2 +-
 17 files changed, 571 insertions(+), 82 deletions(-)

diff --git a/README.md b/README.md
index 68e3440..1a2c886 100644
--- a/README.md
+++ b/README.md
@@ -48,7 +48,7 @@
 
 ## To-Dos
 
-- [ ] [代码翻译](https://github.com/krahets/hello-algo/issues/15)（JavaScript, TypeScript, C, C#, ... 请求大佬帮助）
+- [x] [代码翻译](https://github.com/krahets/hello-algo/issues/15)（JavaScript, TypeScript, C, C#, ... 请求大佬帮助）
 - [ ] 数据结构：散列表、堆（优先队列）、图
 - [ ] 算法：搜索与回溯、选择 / 堆排序、动态规划、贪心、分治
 
diff --git a/codes/go/chapter_computational_complexity/leetcode_two_sum.go b/codes/go/chapter_computational_complexity/leetcode_two_sum.go
index c3c34f9..6712d97 100644
--- a/codes/go/chapter_computational_complexity/leetcode_two_sum.go
+++ b/codes/go/chapter_computational_complexity/leetcode_two_sum.go
@@ -7,6 +7,7 @@ package chapter_computational_complexity
 // twoSumBruteForce
 func twoSumBruteForce(nums []int, target int) []int {
 	size := len(nums)
+	// 两层循环，时间复杂度 O(n^2)
 	for i := 0; i < size-1; i++ {
 		for j := i + 1; i < size; j++ {
 			if nums[i]+nums[j] == target {
@@ -19,7 +20,9 @@ func twoSumBruteForce(nums []int, target int) []int {
 
 // twoSumHashTable
 func twoSumHashTable(nums []int, target int) []int {
+	// 辅助哈希表，空间复杂度 O(n)
 	hashTable := map[int]int{}
+	// 单层循环，时间复杂度 O(n)
 	for idx, val := range nums {
 		if preIdx, ok := hashTable[target-val]; ok {
 			return []int{preIdx, idx}
diff --git a/codes/java/chapter_computational_complexity/leetcode_two_sum.java b/codes/java/chapter_computational_complexity/leetcode_two_sum.java
index 328e1c2..a05c014 100644
--- a/codes/java/chapter_computational_complexity/leetcode_two_sum.java
+++ b/codes/java/chapter_computational_complexity/leetcode_two_sum.java
@@ -8,9 +8,10 @@ package chapter_computational_complexity;
 
 import java.util.*;
 
-class solution_brute_force {
+class SolutionBruteForce {
     public int[] twoSum(int[] nums, int target) {
         int size = nums.length;
+        // 两层循环，时间复杂度 O(n^2)
         for (int i = 0; i < size - 1; i++) {
             for (int j = i + 1; j < size; j++) {
                 if (nums[i] + nums[j] == target)
@@ -21,10 +22,12 @@ class solution_brute_force {
     }
 }
 
-class solution_hash_map {
+class SolutionHashMap {
     public int[] twoSum(int[] nums, int target) {
         int size = nums.length;
+        // 辅助哈希表，空间复杂度 O(n)
         Map<Integer, Integer> dic = new HashMap<>();
+        // 单层循环，时间复杂度 O(n)
         for (int i = 0; i < size; i++) {
             if (dic.containsKey(target - nums[i])) {
                 return new int[] { dic.get(target - nums[i]), i };
@@ -43,11 +46,11 @@ public class leetcode_two_sum {
         
         // ====== Driver Code ======
         // 方法一
-        solution_brute_force slt1 = new solution_brute_force();
+        SolutionBruteForce slt1 = new SolutionBruteForce();
         int[] res = slt1.twoSum(nums, target);
         System.out.println(Arrays.toString(res));
         // 方法二
-        solution_hash_map slt2 = new solution_hash_map();
+        SolutionHashMap slt2 = new SolutionHashMap();
         res = slt2.twoSum(nums, target);
         System.out.println(Arrays.toString(res));
     }
diff --git a/codes/java/chapter_computational_complexity/space_complexity_types.java b/codes/java/chapter_computational_complexity/space_complexity_types.java
index 182dc47..0a8e1bd 100644
--- a/codes/java/chapter_computational_complexity/space_complexity_types.java
+++ b/codes/java/chapter_computational_complexity/space_complexity_types.java
@@ -100,7 +100,7 @@ public class space_complexity_types {
         quadratic(n);
         quadraticRecur(n);
         // 指数阶
-        TreeNode tree = buildTree(n);
-        PrintUtil.printTree(tree);
+        TreeNode root = buildTree(n);
+        PrintUtil.printTree(root);
     }
 }
diff --git a/codes/java/chapter_computational_complexity/time_complexity_types.java b/codes/java/chapter_computational_complexity/time_complexity_types.java
index 2237ed9..27dc8ca 100644
--- a/codes/java/chapter_computational_complexity/time_complexity_types.java
+++ b/codes/java/chapter_computational_complexity/time_complexity_types.java
@@ -29,7 +29,6 @@ public class time_complexity_types {
         int count = 0;
         // 循环次数与数组长度成正比
         for (int num : nums) {
-            // System.out.println(num);
             count++;
         }
         return count;
@@ -38,6 +37,7 @@ public class time_complexity_types {
     /* 平方阶 */
     static int quadratic(int n) {
         int count = 0;
+        // 循环次数与数组长度成平方关系
         for (int i = 0; i < n; i++) {
             for (int j = 0; j < n; j++) {
                 count++;
@@ -47,18 +47,22 @@ public class time_complexity_types {
     }
 
     /* 平方阶（冒泡排序） */
-    static void bubbleSort(int[] nums) {
-        int n = nums.length;
-        for (int i = 0; i < n - 1; i++) {
-            for (int j = 0; j < n - 1 - i; j++) {
+    static int bubbleSort(int[] nums) {
+        int count = 0;  // 计数器
+        // 外循环：待排序元素数量为 n-1, n-2, ..., 1
+        for (int i = nums.length - 1; i > 0; i--) {
+            // 内循环：冒泡操作
+            for (int j = 0; j < i; j++) {
                 if (nums[j] > nums[j + 1]) {
-                    // 交换 nums[j] 和 nums[j + 1]
+                    // 交换 nums[j] 与 nums[j + 1]
                     int tmp = nums[j];
                     nums[j] = nums[j + 1];
                     nums[j + 1] = tmp;
+                    count += 3;  // 元素交换包含 3 个单元操作
                 }
             }
         }
+        return count;
     }
 
     /* 指数阶（循环实现） */
@@ -135,6 +139,11 @@ public class time_complexity_types {
 
         count = quadratic(n);
         System.out.println("平方阶的计算操作数量 = " + count);
+        int[] nums = new int[n];
+        for (int i = 0; i < n; i++)
+            nums[i] = n - i;  // [n,n-1,...,2,1]
+        count = bubbleSort(nums);
+        System.out.println("平方阶（冒泡排序）的计算操作数量 = " + count);
 
         count = exponential(n);
         System.out.println("指数阶（循环实现）的计算操作数量 = " + count);
diff --git a/codes/python/chapter_array_and_linkedlist/array.py b/codes/python/chapter_array_and_linkedlist/array.py
index f505aa4..826c3bd 100644
--- a/codes/python/chapter_array_and_linkedlist/array.py
+++ b/codes/python/chapter_array_and_linkedlist/array.py
@@ -57,6 +57,7 @@ def find(nums, target):
             return i
     return -1
 
+
 """ Driver Code """
 if __name__ == "__main__":
     """ 初始化数组 """
diff --git a/codes/python/chapter_array_and_linkedlist/linked_list.py b/codes/python/chapter_array_and_linkedlist/linked_list.py
index 1f5aaca..d176d8d 100644
--- a/codes/python/chapter_array_and_linkedlist/linked_list.py
+++ b/codes/python/chapter_array_and_linkedlist/linked_list.py
@@ -41,9 +41,8 @@ def find(head, target):
         index += 1
     return -1
 
-"""
-Driver Code
-"""
+
+""" Driver Code """
 if __name__ == "__main__":
     """ 初始化链表 """
     # 初始化各个结点 
diff --git a/codes/python/chapter_array_and_linkedlist/list.py b/codes/python/chapter_array_and_linkedlist/list.py
index 8f45765..4cd9b03 100644
--- a/codes/python/chapter_array_and_linkedlist/list.py
+++ b/codes/python/chapter_array_and_linkedlist/list.py
@@ -8,9 +8,8 @@ import sys, os.path as osp
 sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
 from include import *
 
-"""
-Driver Code
-"""
+
+""" Driver Code """
 if __name__ == "__main__":
     """ 初始化列表 """
     list = [1, 3, 2, 5, 4]
diff --git a/codes/python/chapter_array_and_linkedlist/my_list.py b/codes/python/chapter_array_and_linkedlist/my_list.py
index e5202ed..54a75b3 100644
--- a/codes/python/chapter_array_and_linkedlist/my_list.py
+++ b/codes/python/chapter_array_and_linkedlist/my_list.py
@@ -8,7 +8,6 @@ import sys, os.path as osp
 sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
 from include import *
 
-
 """ 列表类简易实现 """
 class MyList:
     """ 构造函数 """
@@ -71,9 +70,7 @@ class MyList:
         return self._nums[:self._size]
 
 
-"""
-Driver Code
-"""
+""" Driver Code """
 if __name__ == "__main__":
     """ 初始化列表 """
     list = MyList()
diff --git a/codes/python/chapter_computational_complexity/leetcode_two_sum.py b/codes/python/chapter_computational_complexity/leetcode_two_sum.py
index f235ddb..61b7b5c 100644
--- a/codes/python/chapter_computational_complexity/leetcode_two_sum.py
+++ b/codes/python/chapter_computational_complexity/leetcode_two_sum.py
@@ -8,3 +8,34 @@ import sys, os.path as osp
 sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
 from include import *
 
+class SolutionBruteForce:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        for i in range(len(nums) - 1):
+            for j in range(i + 1, len(nums)):
+                if nums[i] + nums[j] == target:
+                    return i, j
+        return []
+
+class SolutionHashMap:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        dic = {}
+        for i in range(len(nums)):
+            if target - nums[i] in dic:
+                return dic[target - nums[i]], i
+            dic[nums[i]] = i
+        return []
+
+
+""" Driver Code """
+if __name__ == '__main__':
+    # ======= Test Case =======
+    nums = [ 2,7,11,15 ];
+    target = 9;
+    
+    # ====== Driver Code ======
+    # 方法一
+    res = SolutionBruteForce().twoSum(nums, target);
+    print(res)
+    # 方法二
+    res = SolutionHashMap().twoSum(nums, target);
+    print(res)
diff --git a/codes/python/chapter_computational_complexity/space_complexity_types.py b/codes/python/chapter_computational_complexity/space_complexity_types.py
index 8905c5b..e8568a2 100644
--- a/codes/python/chapter_computational_complexity/space_complexity_types.py
+++ b/codes/python/chapter_computational_complexity/space_complexity_types.py
@@ -8,3 +8,71 @@ import sys, os.path as osp
 sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
 from include import *
 
+""" 函数 """
+def function():
+    # do something
+    return 0
+
+""" 常数阶 """
+def constant(n):
+    # 常量、变量、对象占用 O(1) 空间
+    a = 0
+    nums = [0] * 10000
+    node = ListNode(0)
+    # 循环中的变量占用 O(1) 空间
+    for _ in range(n):
+         c = 0
+    # 循环中的函数占用 O(1) 空间
+    for _ in range(n):
+        function()
+
+""" 线性阶 """
+def linear(n):
+    # 长度为 n 的列表占用 O(n) 空间
+    nums = [0] * n
+    # 长度为 n 的哈希表占用 O(n) 空间
+    mapp = {}
+    for i in range(n):
+        mapp[i] = str(i)
+
+""" 线性阶（递归实现） """
+def linearRecur(n):
+    print("递归 n = ", n)
+    if n == 1: return
+    linearRecur(n - 1)
+
+""" 平方阶 """
+def quadratic(n):
+    # 二维列表占用 O(n^2) 空间
+    num_matrix = [[0] * n for _ in range(n)]
+
+""" 平方阶（递归实现） """
+def quadratic_recur(n):
+    if n <= 0: return 0
+    nums = [0] * n
+    print("递归 n = {} 中的 nums 长度 = {}".format(n, len(nums)))
+    return quadratic_recur(n - 1)
+
+""" 指数阶（建立满二叉树） """
+def build_tree(n):
+    if n == 0: return None
+    root = TreeNode(0)
+    root.left = build_tree(n - 1)
+    root.right = build_tree(n - 1)
+    return root
+
+
+""" Driver Code """
+if __name__ == "__main__":
+    n = 5
+    # 常数阶
+    constant(n)
+    # 线性阶
+    linear(n)
+    linearRecur(n)
+    # 平方阶
+    quadratic(n)
+    quadratic_recur(n)
+    # 指数阶
+    root = build_tree(n)
+    print_tree(root)
diff --git a/codes/python/chapter_computational_complexity/time_complexity_types.py b/codes/python/chapter_computational_complexity/time_complexity_types.py
index e60ae4c..c56ef7b 100644
--- a/codes/python/chapter_computational_complexity/time_complexity_types.py
+++ b/codes/python/chapter_computational_complexity/time_complexity_types.py
@@ -8,3 +8,133 @@ import sys, os.path as osp
 sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
 from include import *
 
+""" 常数阶 """
+def constant(n):
+    count = 0
+    size = 100000
+    for _ in range(size):
+        count += 1
+    return count
+
+""" 线性阶 """
+def linear(n):
+    count = 0
+    for _ in range(n):
+        count += 1
+    return count
+
+""" 线性阶（遍历数组）"""
+def array_traversal(nums):
+    count = 0
+    # 循环次数与数组长度成正比
+    for num in nums:
+        count += 1
+    return count
+
+""" 平方阶 """
+def quadratic(n):
+    count = 0
+    # 循环次数与数组长度成平方关系
+    for i in range(n):
+        for j in range(n):
+            count += 1
+    return count
+
+""" 平方阶（冒泡排序）"""
+def bubble_sort(nums):
+    count = 0  # 计数器
+    # 外循环：待排序元素数量为 n-1, n-2, ..., 1
+    for i in range(len(nums) - 1, 0, -1):
+        # 内循环：冒泡操作
+        for j in range(i):
+            if nums[j] > nums[j + 1]:
+                # 交换 nums[j] 与 nums[j + 1]
+                tmp = nums[j]
+                nums[j] = nums[j + 1]
+                nums[j + 1] = tmp
+                count += 3  # 元素交换包含 3 个单元操作
+    return count
+
+""" 指数阶（循环实现）"""
+def exponential(n):
+    count, base = 0, 1
+    # cell 每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
+    for _ in range(n):
+        for _ in range(base):
+            count += 1
+        base *= 2
+    # count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
+    return count
+
+""" 指数阶（递归实现）"""
+def exp_recur(n):
+    if n == 1: return 1
+    return exp_recur(n - 1) + exp_recur(n - 1) + 1
+
+""" 对数阶（循环实现）"""
+def logarithmic(n):
+    count = 0
+    while n > 1:
+        n = n / 2
+        count += 1
+    return count
+
+""" 对数阶（递归实现）"""
+def log_recur(n):
+    if n <= 1: return 0
+    return log_recur(n / 2) + 1
+
+""" 线性对数阶 """
+def linear_log_recur(n):
+    if n <= 1: return 1
+    count = linear_log_recur(n // 2) + \
+            linear_log_recur(n // 2)
+    for _ in range(n):
+        count += 1
+    return count
+
+""" 阶乘阶（递归实现）"""
+def factorial_recur(n):
+    if n == 0: return 1
+    count = 0
+    # 从 1 个分裂出 n 个
+    for _ in range(n):
+        count += factorial_recur(n - 1)
+    return count
+
+
+""" Driver Code """
+if __name__ == "__main__":
+    # 可以修改 n 运行，体会一下各种复杂度的操作数量变化趋势
+    n = 8
+    print("输入数据大小 n =", n)
+
+    count = constant(n)
+    print("常数阶的计算操作数量 =", count)
+
+    count = linear(n)
+    print("线性阶的计算操作数量 =", count)
+    count = array_traversal([0] * n)
+    print("线性阶（遍历数组）的计算操作数量 =", count)
+
+    count = quadratic(n)
+    print("平方阶的计算操作数量 =", count)
+    nums = [i for i in range(n, 0, -1)]  # [n,n-1,...,2,1]
+    count = bubble_sort(nums)
+    print("平方阶（冒泡排序）的计算操作数量 =", count)
+
+    count = exponential(n)
+    print("指数阶（循环实现）的计算操作数量 =", count)
+    count = exp_recur(n)
+    print("指数阶（递归实现）的计算操作数量 =", count)
+
+    count = logarithmic(n)
+    print("对数阶（循环实现）的计算操作数量 =", count)
+    count = log_recur(n)
+    print("对数阶（递归实现）的计算操作数量 =", count)
+
+    count = linear_log_recur(n)
+    print("线性对数阶（递归实现）的计算操作数量 =", count)
+
+    count = factorial_recur(n)
+    print("阶乘阶（递归实现）的计算操作数量 =", count)
diff --git a/codes/python/chapter_computational_complexity/worst_best_time_complexity.py b/codes/python/chapter_computational_complexity/worst_best_time_complexity.py
index 7a4b246..7b052a3 100644
--- a/codes/python/chapter_computational_complexity/worst_best_time_complexity.py
+++ b/codes/python/chapter_computational_complexity/worst_best_time_complexity.py
@@ -8,3 +8,27 @@ import sys, os.path as osp
 sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
 from include import *
 
+""" 生成一个数组，元素为: 1, 2, ..., n ，顺序被打乱 """
+def random_numbers(n):
+    # 生成数组 nums =: 1, 2, 3, ..., n 
+    nums = [i for i in range(1, n + 1)]
+    # 随机打乱数组元素
+    random.shuffle(nums)
+    return nums
+
+""" 查找数组 nums 中数字 1 所在索引 """
+def find_one(nums):
+    for i in range(len(nums)):
+        if nums[i] == 1:
+            return i
+    return -1
+
+
+""" Driver Code """
+if __name__ == "__main__":
+    for i in range(10):
+        n = 100
+        nums = random_numbers(n)
+        index = find_one(nums)
+        print("\n数组 [ 1, 2, ..., n ] 被打乱后 =", nums)
+        print("数字 1 的索引为", index)
diff --git a/docs/chapter_computational_complexity/space_complexity.md b/docs/chapter_computational_complexity/space_complexity.md
index b7856b6..c1b3a98 100644
--- a/docs/chapter_computational_complexity/space_complexity.md
+++ b/docs/chapter_computational_complexity/space_complexity.md
@@ -62,7 +62,23 @@ comments: true
 === "Python"
 
     ```python title=""
-    
+    """ 类 """
+    class Node:
+        def __init__(self, x):
+            self.val = x      # 结点值
+            self.next = None  # 指向下一结点的指针（引用）
+
+    """ 函数（或称方法） """
+    def function():
+        # do something...
+        return 0
+
+    def algorithm(n):     # 输入数据
+        a = 0             # 暂存数据（常量）
+        b = 0             # 暂存数据（变量）
+        node = Node(0)    # 暂存数据（对象）
+        c = function()    # 栈帧空间（调用函数）
+        return a + b + c  # 输出数据
     ```
 
 ## 推算方法
@@ -94,7 +110,11 @@ comments: true
 === "Python"
 
     ```python title=""
-    
+    def algorithm(n):
+        a = 0               # O(1)
+        b = [0] * 10000     # O(1)
+        if n > 10:
+            nums = [0] * n  # O(n)
     ```
 
 **在递归函数中，需要注意统计栈帧空间。** 例如函数 `loop()`，在循环中调用了 $n$ 次 `function()` ，每轮中的 `function()` 都返回并释放了栈帧空间，因此空间复杂度仍为 $O(1)$ 。而递归函数 `recur()` 在运行中会同时存在 $n$ 个未返回的 `recur()` ，从而使用 $O(n)$ 的栈帧空间。
@@ -106,13 +126,13 @@ comments: true
         // do something
         return 0;
     }
-    /* 循环 */
+    /* 循环 O(1) */
     void loop(int n) {
         for (int i = 0; i < n; i++) {
             function();
         }
     }
-    /* 递归 */
+    /* 递归 O(n) */
     void recur(int n) {
         if (n == 1) return;
         return recur(n - 1);
@@ -128,7 +148,19 @@ comments: true
 === "Python"
 
     ```python title=""
-    
+    def function():
+        # do something
+        return 0
+
+    """ 循环 O(1) """
+    def loop(n):
+        for _ in range(n):
+            function()
+
+    """ 递归 O(n) """
+    def recur(n):
+        if n == 1: return
+        return recur(n - 1)
     ```
 
 ## 常见类型
@@ -186,7 +218,18 @@ $$
 === "Python"
 
     ```python title="space_complexity_types.py"
-    
+    """ 常数阶 """
+    def constant(n):
+        # 常量、变量、对象占用 O(1) 空间
+        a = 0
+        nums = [0] * 10000
+        node = ListNode(0)
+        # 循环中的变量占用 O(1) 空间
+        for _ in range(n):
+            c = 0
+        # 循环中的函数占用 O(1) 空间
+        for _ in range(n):
+            function()
     ```
 
 ### 线性阶 $O(n)$
@@ -222,7 +265,14 @@ $$
 === "Python"
 
     ```python title="space_complexity_types.py"
-    
+    """ 线性阶 """
+    def linear(n):
+        # 长度为 n 的列表占用 O(n) 空间
+        nums = [0] * n
+        # 长度为 n 的哈希表占用 O(n) 空间
+        mapp = {}
+        for i in range(n):
+            mapp[i] = str(i)
     ```
 
 以下递归函数会同时存在 $n$ 个未返回的 `algorithm()` 函数，使用 $O(n)$ 大小的栈帧空间。
@@ -247,7 +297,11 @@ $$
 === "Python"
 
     ```python title="space_complexity_types.py"
-    
+    """ 线性阶（递归实现） """
+    def linearRecur(n):
+        print("递归 n = ", n)
+        if n == 1: return
+        linearRecur(n - 1)
     ```
 
 ![space_complexity_recursive_linear](space_complexity.assets/space_complexity_recursive_linear.png)
@@ -286,7 +340,10 @@ $$
 === "Python"
 
     ```python title="space_complexity_types.py"
-    
+    """ 平方阶 """
+    def quadratic(n):
+        # 二维列表占用 O(n^2) 空间
+        num_matrix = [[0] * n for _ in range(n)]
     ```
 
 在以下递归函数中，同时存在 $n$ 个未返回的 `algorihtm()` ，并且每个函数中都初始化了一个数组，长度分别为 $n, n-1, n-2, ..., 2, 1$ ，平均长度为 $\frac{n}{2}$ ，因此总体使用 $O(n^2)$ 空间。
@@ -297,8 +354,8 @@ $$
     /* 平方阶（递归实现） */
     int quadraticRecur(int n) {
         if (n <= 0) return 0;
+        // 数组 nums 长度为 n, n-1, ..., 2, 1
         int[] nums = new int[n];
-        System.out.println("递归 n = " + n + " 中的 nums 长度 = " + nums.length);
         return quadraticRecur(n - 1);
     }
     ```
@@ -312,7 +369,12 @@ $$
 === "Python"
 
     ```python title="space_complexity_types.py"
-    
+    """ 平方阶（递归实现） """
+    def quadratic_recur(n):
+        if n <= 0: return 0
+        # 数组 nums 长度为 n, n-1, ..., 2, 1
+        nums = [0] * n
+        return quadratic_recur(n - 1)
     ```
 
 ![space_complexity_recursive_quadratic](space_complexity.assets/space_complexity_recursive_quadratic.png)
@@ -345,7 +407,13 @@ $$
 === "Python"
 
     ```python title="space_complexity_types.py"
-
+    """ 指数阶（建立满二叉树） """
+    def build_tree(n):
+        if n == 0: return None
+        root = TreeNode(0)
+        root.left = build_tree(n - 1)
+        root.right = build_tree(n - 1)
+        return root
     ```
 
 ![space_complexity_exponential](space_complexity.assets/space_complexity_exponential.png)
diff --git a/docs/chapter_computational_complexity/space_time_tradeoff.md b/docs/chapter_computational_complexity/space_time_tradeoff.md
index 0f82f6c..07975a5 100644
--- a/docs/chapter_computational_complexity/space_time_tradeoff.md
+++ b/docs/chapter_computational_complexity/space_time_tradeoff.md
@@ -17,16 +17,18 @@
 === "Java"
 
     ```java title="" title="leetcode_two_sum.java"
-    public int[] twoSum(int[] nums, int target) {
-        int size = nums.length;
-        // 外层 * 内层循环，时间复杂度为 O(n)
-        for (int i = 0; i < size - 1; i++) {
-            for (int j = i + 1; j < size; j++) {
-                if (nums[i] + nums[j] == target)
-                    return new int[] { i, j };
+    class SolutionBruteForce {
+        public int[] twoSum(int[] nums, int target) {
+            int size = nums.length;
+            // 两层循环，时间复杂度 O(n^2)
+            for (int i = 0; i < size - 1; i++) {
+                for (int j = i + 1; j < size; j++) {
+                    if (nums[i] + nums[j] == target)
+                        return new int[] { i, j };
+                }
             }
+            return new int[0];
         }
-        return new int[0];
     }
     ```
 
@@ -39,14 +41,22 @@
 === "Python"
 
     ```python title="leetcode_two_sum.py"
-    
+    class SolutionBruteForce:
+        def twoSum(self, nums: List[int], target: int) -> List[int]:
+            # 两层循环，时间复杂度 O(n^2)
+            for i in range(len(nums) - 1):
+                for j in range(i + 1, len(nums)):
+                    if nums[i] + nums[j] == target:
+                        return i, j
+            return []
     ```
 
 === "Go"
 
     ```go title="leetcode_two_sum.go"
-    func twoSum(nums []int, target int) []int {
+    func twoSumBruteForce(nums []int, target int) []int {
         size := len(nums)
+        // 两层循环，时间复杂度 O(n^2)
         for i := 0; i < size-1; i++ {
             for j := i + 1; i < size; j++ {
                 if nums[i]+nums[j] == target {
@@ -58,8 +68,6 @@
     }
     ```
 
-
-
 ### 方法二：辅助哈希表
 
 时间复杂度 $O(N)$ ，空间复杂度 $O(N)$ ，属于「空间换时间」。
@@ -69,18 +77,20 @@
 === "Java"
 
     ```java title="" title="leetcode_two_sum.java"
-    public int[] twoSum(int[] nums, int target) {
-        int size = nums.length;
-        // 辅助哈希表，空间复杂度 O(n)
-        Map<Integer, Integer> dic = new HashMap<>();
-        // 单层循环，时间复杂度 O(n)
-        for (int i = 0; i < size; i++) {
-            if (dic.containsKey(target - nums[i])) {
-                return new int[] { dic.get(target - nums[i]), i };
+    class SolutionHashMap {
+        public int[] twoSum(int[] nums, int target) {
+            int size = nums.length;
+            // 辅助哈希表，空间复杂度 O(n)
+            Map<Integer, Integer> dic = new HashMap<>();
+            // 单层循环，时间复杂度 O(n)
+            for (int i = 0; i < size; i++) {
+                if (dic.containsKey(target - nums[i])) {
+                    return new int[] { dic.get(target - nums[i]), i };
+                }
+                dic.put(nums[i], i);
             }
-            dic.put(nums[i], i);
+            return new int[0];
         }
-        return new int[0];
     }
     ```
 
@@ -93,14 +103,25 @@
 === "Python"
 
     ```python title="leetcode_two_sum.py"
-    
+    class SolutionHashMap:
+        def twoSum(self, nums: List[int], target: int) -> List[int]:
+            # 辅助哈希表，空间复杂度 O(n)
+            dic = {}
+            # 单层循环，时间复杂度 O(n)
+            for i in range(len(nums)):
+                if target - nums[i] in dic:
+                    return dic[target - nums[i]], i
+                dic[nums[i]] = i
+            return []
     ```
 
 === "Go"
 
     ```go title="leetcode_two_sum.go"
     func twoSumHashTable(nums []int, target int) []int {
+        // 辅助哈希表，空间复杂度 O(n)
         hashTable := map[int]int{}
+        // 单层循环，时间复杂度 O(n)
         for idx, val := range nums {
             if preIdx, ok := hashTable[target-val]; ok {
                 return []int{preIdx, idx}
diff --git a/docs/chapter_computational_complexity/time_complexity.md b/docs/chapter_computational_complexity/time_complexity.md
index 7742aec..a25cabf 100644
--- a/docs/chapter_computational_complexity/time_complexity.md
+++ b/docs/chapter_computational_complexity/time_complexity.md
@@ -42,7 +42,14 @@ $$
 === "Python"
 
     ```python title=""
-    
+    # 在某运行平台下
+    def algorithm(n):
+        a = 2      # 1 ns
+        a = a + 1  # 1 ns
+        a = a * 2  # 10 ns
+        # 循环 n 次
+        for _ in range(n):  # 1 ns
+            print(0)        # 5 ns
     ```
 
 但实际上， **统计算法的运行时间既不合理也不现实。** 首先，我们不希望预估时间和运行平台绑定，毕竟算法需要跑在各式各样的平台之上。其次，我们很难获知每一种操作的运行时间，这为预估过程带来了极大的难度。
@@ -87,7 +94,17 @@ $$
 === "Python"
 
     ```python title=""
-    
+    # 算法 A 时间复杂度：常数阶
+    def algorithm_A(n):
+        print(0)
+    # 算法 B 时间复杂度：线性阶
+    def algorithm_B(n):
+        for _ in range(n):
+            print(0)
+    # 算法 C 时间复杂度：常数阶
+    def algorithm_C(n):
+        for _ in range(1000000):
+            print(0)
     ```
 
 ![time_complexity_first_example](time_complexity.assets/time_complexity_first_example.png)
@@ -105,9 +122,11 @@ $$
 ## 函数渐进上界
 
 设算法「计算操作数量」为 $T(n)$  ，其是一个关于输入数据大小 $n$ 的函数。例如，以下算法的操作数量为
+
 $$
 T(n) = 3 + 2n
 $$
+
 === "Java"
 
     ```java title=""
@@ -131,14 +150,21 @@ $$
 === "Python"
 
     ```python title=""
-    
+    def algorithm(n):
+        a = 1  # +1
+        a = a + 1  # +1
+        a = a * 2  # +1
+        # 循环 n 次
+        for i in range(n):  # +1
+            print(0)        # +1
+    }
     ```
 
 $T(n)$ 是个一次函数，说明时间增长趋势是线性的，因此易得时间复杂度是线性阶。
 
 我们将线性阶的时间复杂度记为 $O(n)$ ，这个数学符号被称为「大 $O$ 记号 Big-$O$ Notation」，代表函数 $T(n)$ 的「渐进上界 asymptotic upper bound」。
 
-我们要推算时间复杂度，本质上是在计算「操作数量函数 $T(n)$ 」的渐进上界。下面我们先来看看函数渐进上界的数学定义。 
+我们要推算时间复杂度，本质上是在计算「操作数量函数 $T(n)$ 」的渐进上界。下面我们先来看看函数渐进上界的数学定义。
 
 !!! abstract "函数渐进上界"
 
@@ -174,6 +200,7 @@ $T(n)$ 是个一次函数，说明时间增长趋势是线性的，因此易得
 3. **循环嵌套时使用乘法。** 总操作数量等于外层循环和内层循环操作数量之积，每一层循环依然可以分别套用上述 `1.` 和 `2.` 技巧。
 
 根据以下示例，使用上述技巧前、后的统计结果分别为
+
 $$
 \begin{aligned}
 T(n) & = 2n(n + 1) + (5n + 1) + 2 & \text{完整统计 (-.-|||)} \newline
@@ -181,6 +208,7 @@ T(n) & = 2n(n + 1) + (5n + 1) + 2 & \text{完整统计 (-.-|||)} \newline
 T(n) & = n^2 + n & \text{偷懒统计 (o.O)}
 \end{aligned}
 $$
+
 最终，两者都能推出相同的时间复杂度结果，即 $O(n^2)$ 。
 
 === "Java"
@@ -211,7 +239,16 @@ $$
 === "Python"
 
     ```python title=""
-    
+    def algorithm(n):
+        a = 1      # +0（技巧 1）
+        a = a + n  # +0（技巧 1）
+        # +n（技巧 2）
+        for i in range(5 * n + 1):
+            print(0)
+        # +n*n（技巧 3）
+        for i in range(2 * n):
+            for j in range(n + 1):
+                print(0)
     ```
 
 ### 2. 判断渐进上界
@@ -279,7 +316,13 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 常数阶 """
+    def constant(n):
+        count = 0
+        size = 100000
+        for _ in range(size):
+            count += 1
+        return count
     ```
 
 ### 线性阶 $O(n)$
@@ -307,7 +350,12 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 线性阶 """
+    def linear(n):
+        count = 0
+        for _ in range(n):
+            count += 1
+        return count
     ```
 
 「遍历数组」和「遍历链表」等操作，时间复杂度都为 $O(n)$ ，其中 $n$ 为数组或链表的长度。
@@ -339,7 +387,13 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 线性阶（遍历数组）"""
+    def array_traversal(nums):
+        count = 0
+        # 循环次数与数组长度成正比
+        for num in nums:
+            count += 1
+        return count
     ```
 
 ### 平方阶 $O(n^2)$
@@ -352,6 +406,7 @@ $$
     /* 平方阶 */
     int quadratic(int n) {
         int count = 0;
+        // 循环次数与数组长度成平方关系
         for (int i = 0; i < n; i++) {
             for (int j = 0; j < n; j++) {
                 count++;
@@ -370,7 +425,14 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 平方阶 """
+    def quadratic(n):
+        count = 0
+        # 循环次数与数组长度成平方关系
+        for i in range(n):
+            for j in range(n):
+                count += 1
+        return count
     ```
 
 ![time_complexity_constant_linear_quadratic](time_complexity.assets/time_complexity_constant_linear_quadratic.png)
@@ -387,18 +449,22 @@ $$
 
     ```java title="" title="time_complexity_types.java"
     /* 平方阶（冒泡排序） */
-    void bubbleSort(int[] nums) {
-        int n = nums.length;
-        for (int i = 0; i < n - 1; i++) {
-            for (int j = 0; j < n - 1 - i; j++) {
+    int bubbleSort(int[] nums) {
+        int count = 0;  // 计数器
+        // 外循环：待排序元素数量为 n-1, n-2, ..., 1
+        for (int i = nums.length - 1; i > 0; i--) {
+            // 内循环：冒泡操作
+            for (int j = 0; j < i; j++) {
                 if (nums[j] > nums[j + 1]) {
-                    // 交换 nums[j] 和 nums[j + 1]
+                    // 交换 nums[j] 与 nums[j + 1]
                     int tmp = nums[j];
                     nums[j] = nums[j + 1];
                     nums[j + 1] = tmp;
+                    count += 3;  // 元素交换包含 3 个单元操作
                 }
             }
         }
+        return count;
     }
     ```
 
@@ -411,7 +477,20 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 平方阶（冒泡排序）"""
+    def bubble_sort(nums):
+        count = 0  # 计数器
+        # 外循环：待排序元素数量为 n-1, n-2, ..., 1
+        for i in range(len(nums) - 1, 0, -1):
+            # 内循环：冒泡操作
+            for j in range(i):
+                if nums[j] > nums[j + 1]:
+                    # 交换 nums[j] 与 nums[j + 1]
+                    tmp = nums[j]
+                    nums[j] = nums[j + 1]
+                    nums[j + 1] = tmp
+                    count += 3  # 元素交换包含 3 个单元操作
+        return count
     ```
 
 ### 指数阶 $O(2^n)$
@@ -425,7 +504,7 @@ $$
 === "Java"
 
     ```java title="" title="time_complexity_types.java"
-    /* 指数阶（遍历实现） */
+    /* 指数阶（循环实现） */
     int exponential(int n) {
         int count = 0, base = 1;
         // cell 每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
@@ -449,7 +528,16 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 指数阶（循环实现）"""
+    def exponential(n):
+        count, base = 0, 1
+        # cell 每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
+        for _ in range(n):
+            for _ in range(base):
+                count += 1
+            base *= 2
+        # count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
+        return count
     ```
 
 ![time_complexity_exponential](time_complexity.assets/time_complexity_exponential.png)
@@ -477,7 +565,10 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 指数阶（递归实现）"""
+    def exp_recur(n):
+        if n == 1: return 1
+        return exp_recur(n - 1) + exp_recur(n - 1) + 1
     ```
 
 ### 对数阶 $O(\log n)$
@@ -511,7 +602,13 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 对数阶（循环实现）"""
+    def logarithmic(n):
+        count = 0
+        while n > 1:
+            n = n / 2
+            count += 1
+        return count
     ```
 
 ![time_complexity_logarithmic](time_complexity.assets/time_complexity_logarithmic.png)
@@ -539,7 +636,10 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 对数阶（递归实现）"""
+    def log_recur(n):
+        if n <= 1: return 0
+        return log_recur(n / 2) + 1
     ```
 
 ### 线性对数阶 $O(n \log n)$
@@ -572,7 +672,14 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 线性对数阶 """
+    def linear_log_recur(n):
+        if n <= 1: return 1
+        count = linear_log_recur(n // 2) + \
+                linear_log_recur(n // 2)
+        for _ in range(n):
+            count += 1
+        return count
     ```
 
 ![time_complexity_logarithmic_linear](time_complexity.assets/time_complexity_logarithmic_linear.png)
@@ -613,7 +720,14 @@ $$
 === "Python"
 
     ```python title="time_complexity_types.py"
-    
+    """ 阶乘阶（递归实现）"""
+    def factorial_recur(n):
+        if n == 0: return 1
+        count = 0
+        # 从 1 个分裂出 n 个
+        for _ in range(n):
+            count += factorial_recur(n - 1)
+        return count
     ```
 
 ![time_complexity_factorial](time_complexity.assets/time_complexity_factorial.png)
@@ -681,7 +795,29 @@ $$
 === "Python"
 
     ```python title="worst_best_time_complexity.py"
-    
+    """ 生成一个数组，元素为: 1, 2, ..., n ，顺序被打乱 """
+    def random_numbers(n):
+        # 生成数组 nums =: 1, 2, 3, ..., n 
+        nums = [i for i in range(1, n + 1)]
+        # 随机打乱数组元素
+        random.shuffle(nums)
+        return nums
+
+    """ 查找数组 nums 中数字 1 所在索引 """
+    def find_one(nums):
+        for i in range(len(nums)):
+            if nums[i] == 1:
+                return i
+        return -1
+
+    """ Driver Code """
+    if __name__ == "__main__":
+        for i in range(10):
+            n = 100
+            nums = random_numbers(n)
+            index = find_one(nums)
+            print("\n数组 [ 1, 2, ..., n ] 被打乱后 =", nums)
+            print("数字 1 的索引为", index)
     ```
 
 !!! tip
diff --git a/docs/chapter_preface/index.md b/docs/chapter_preface/index.md
index 7c7a9be..597649e 100644
--- a/docs/chapter_preface/index.md
+++ b/docs/chapter_preface/index.md
@@ -111,7 +111,7 @@ comments: true
 
 ## 致谢
 
-感谢本开源书的每一位撰稿人，是他们的无私奉献让这本书变得更好，他们的 GitHub ID（按首次提交时间排序）为：krahets, *（等待下一位创作者）*
+感谢本开源书的每一位撰稿人，是他们的无私奉献让这本书变得更好，他们的 GitHub ID（按首次提交时间排序）为：krahets, Reanon.
 
 本书的成书过程中，我获得了许多人的帮助，包括但不限于：