1. Fix the import error.
2. Some codes fine tuning.
This commit is contained in:
Yudong Jin
parent
0585f20970
commit
9f883d5888
@ -4,11 +4,9 @@ Created Time: 2022-11-25
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Author: timi (xisunyy@163.com)
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'''
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from include import *
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import sys
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import os.path as osp
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import sys, os.path as osp
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sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
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from include import *
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""" 冒泡排序 """
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def bubble_sort(nums):
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@ -17,11 +15,10 @@ def bubble_sort(nums):
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for i in range(n - 1, -1, -1):
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# 内循环：冒泡操作
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for j in range(i):
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# 交换 nums[j] 与 nums[j + 1]
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if nums[j] > nums[j + 1]:
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# 交换 nums[j] 与 nums[j + 1]
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nums[j], nums[j + 1] = nums[j + 1], nums[j]
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""" 冒泡排序（标志优化） """
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def bubble_sort_with_flag(nums):
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n = len(nums)
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@ -30,19 +27,20 @@ def bubble_sort_with_flag(nums):
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flag = False # 初始化标志位
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# 内循环：冒泡操作
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for j in range(i):
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# 交换 nums[j] 与 nums[j + 1]
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if nums[j] > nums[j + 1]:
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# 交换 nums[j] 与 nums[j + 1]
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nums[j], nums[j + 1] = nums[j + 1], nums[j]
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flag = True # 记录交换元素
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if not flag:
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break # 此轮冒泡未交换任何元素，直接跳出
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break # 此轮冒泡未交换任何元素，直接跳出
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""" Driver Code """
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if __name__ == '__main__':
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nums = [4, 1, 3, 1, 5, 2]
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bubble_sort(nums)
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print("排序后数组 nums = ", nums)
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print("排序后数组 nums =", nums)
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nums1 = [4, 1, 3, 1, 5, 2]
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bubble_sort_with_flag(nums1)
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print("排序后数组 nums = ", nums1)
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print("排序后数组 nums =", nums1)
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@ -4,11 +4,9 @@ Created Time: 2022-11-25
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Author: timi (xisunyy@163.com)
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'''
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from include import *
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import sys
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import os.path as osp
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import sys, os.path as osp
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sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
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from include import *
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""" 插入排序 """
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def insertion_sort(nums):
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@ -20,9 +18,10 @@ def insertion_sort(nums):
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while j >= 0 and nums[j] > base:
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nums[j + 1] = nums[j] # 1. 将 nums[j] 向右移动一位
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j -= 1
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nums[j + 1] = base # 2. 将 base 赋值到正确位置
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nums[j + 1] = base # 2. 将 base 赋值到正确位置
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""" Driver Code """
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if __name__ == '__main__':
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nums = [4, 1, 3, 1, 5, 2]
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insertion_sort(nums)
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@ -4,11 +4,9 @@ Created Time: 2022-11-25
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Author: timi (xisunyy@163.com)
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'''
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from include import *
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import sys
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import os.path as osp
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import sys, os.path as osp
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sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
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from include import *
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"""
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合并左子数组和右子数组
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@ -43,16 +41,17 @@ def merge(nums, left, mid, right):
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def merge_sort(nums, left, right):
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# 终止条件
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if left >= right:
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return # 当子数组长度为 1 时终止递归
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return # 当子数组长度为 1 时终止递归
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# 划分阶段
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mid = (left + right) // 2 # 计算中点
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merge_sort(nums, left, mid) # 递归左子数组
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mid = (left + right) // 2 # 计算中点
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merge_sort(nums, left, mid) # 递归左子数组
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merge_sort(nums, mid + 1, right) # 递归右子数组
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# 合并阶段
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merge(nums, left, mid, right)
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""" Driver Code """
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if __name__ == '__main__':
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nums = [4, 1, 3, 1, 5, 2]
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merge_sort(nums, 0, len(nums) - 1)
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print("归并排序完成后 nums = ", nums)
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print("归并排序完成后 nums =", nums)
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@ -4,15 +4,12 @@ Created Time: 2022-11-25
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Author: timi (xisunyy@163.com)
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'''
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from include import *
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import sys
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import os.path as osp
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import sys, os.path as osp
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sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
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from include import *
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""" 快速排序类 """
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class QuickSort(object):
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class QuickSort:
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""" 哨兵划分 """
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def partition(self, nums, left, right):
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# 以 nums[left] 作为基准数
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@ -39,10 +36,8 @@ class QuickSort(object):
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self.quick_sort(nums, left, pivot - 1)
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self.quick_sort(nums, pivot + 1, right)
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""" 快速排序类（中位基准数优化）"""
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class QuickSortMedian():
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class QuickSortMedian:
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""" 选取三个元素的中位数 """
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def median_three(self, nums, left, mid, right):
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# 使用了异或操作来简化代码
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@ -68,7 +63,7 @@ class QuickSortMedian():
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i += 1 # 从左向右找首个大于基准数的元素
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# 元素交换
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nums[i], nums[j] = nums[j], nums[i]
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# 将基准数交换至两子数组的分界线
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# 将基准数交换至两子数组的分界线
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nums[i], nums[left] = nums[left], nums[i]
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return i # 返回基准数的索引
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@ -82,10 +77,8 @@ class QuickSortMedian():
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self.quick_sort(nums, left, pivot - 1)
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self.quick_sort(nums, pivot + 1, right)
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""" 快速排序类（尾递归优化） """
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class QuickSortTailCall():
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class QuickSortTailCall:
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""" 哨兵划分 """
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def partition(self, nums, left, right):
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# 以 nums[left] 作为基准数
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@ -99,7 +92,7 @@ class QuickSortTailCall():
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nums[i], nums[j] = nums[j], nums[i]
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# 将基准数交换至两子数组的分界线
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nums[i], nums[left] = nums[left], nums[i]
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return i # 返回基准数的索引
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return i # 返回基准数的索引
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""" 快速排序（尾递归优化） """
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def quick_sort(self, nums, left, right):
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@ -110,24 +103,25 @@ class QuickSortTailCall():
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# 对两个子数组中较短的那个执行快排
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if pivot - left < right - pivot:
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self.quick_sort(nums, left, pivot - 1) # 递归排序左子数组
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left = pivot + 1 # 剩余待排序区间为 [pivot + 1, right]
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left = pivot + 1 # 剩余待排序区间为 [pivot + 1, right]
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else:
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self.quick_sort(nums, pivot + 1, right) # 递归排序右子数组
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right = pivot - 1 # 剩余待排序区间为 [left, pivot - 1]
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right = pivot - 1 # 剩余待排序区间为 [left, pivot - 1]
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""" Driver Code """
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if __name__ == '__main__':
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# 快速排序
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nums = [4, 1, 3, 1, 5, 2]
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QuickSort().quick_sort(nums, 0, len(nums) - 1)
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print("快速排序完成后 nums = ", nums)
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print("快速排序完成后 nums =", nums)
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# 快速排序（中位基准数优化）
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nums1 = [4, 1, 3, 1, 5, 2]
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QuickSortMedian().quick_sort(nums1, 0, len(nums1) - 1)
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print("快速排序（中位基准数优化）完成后 nums = ", nums)
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print("快速排序（中位基准数优化）完成后 nums =", nums)
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# 快速排序（尾递归优化）
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nums2 = [4, 1, 3, 1, 5, 2]
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QuickSortTailCall().quick_sort(nums, 0, len(nums2) - 1)
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print("快速排序（尾递归优化）完成后 nums = ", nums)
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print("快速排序（尾递归优化）完成后 nums =", nums)
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@ -1,3 +1,7 @@
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---
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comments: true
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---
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# 权衡时间与空间
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理想情况下，我们希望算法的时间复杂度和空间复杂度都能够达到最优，而实际上，同时优化时间复杂度和空间复杂度是非常困难的。
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@ -47,9 +47,7 @@ comments: true
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## 算法流程
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1. 设数组长度为 $n$ ，完成第一轮「冒泡」后，数组最大元素已在正确位置，接下来只需排序剩余 $n - 1$ 个元素。
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2. 同理，对剩余 $n - 1$ 个元素执行「冒泡」，可将第二大元素交换至正确位置，因而待排序元素只剩 $n - 2$ 个。
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3. 以此类推…… **循环 $n - 1$ 轮「冒泡」，即可完成整个数组的排序**。
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@ -76,6 +74,21 @@ comments: true
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}
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```
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=== "Python"
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```python
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""" 冒泡排序 """
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def bubble_sort(nums):
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n = len(nums)
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# 外循环：待排序元素数量为 n-1, n-2, ..., 1
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for i in range(n - 1, -1, -1):
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# 内循环：冒泡操作
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for j in range(i):
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if nums[j] > nums[j + 1]:
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# 交换 nums[j] 与 nums[j + 1]
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nums[j], nums[j + 1] = nums[j + 1], nums[j]
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```
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## 算法特性
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**时间复杂度 $O(n^2)$ ：** 各轮「冒泡」遍历的数组长度为 $n - 1$ , $n - 2$ , $\cdots$ , $2$ , $1$ 次，求和为 $\frac{(n - 1) n}{2}$ ，因此使用 $O(n^2)$ 时间。
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@ -116,3 +129,22 @@ comments: true
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}
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}
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```
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=== "Python"
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```python
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""" 冒泡排序（标志优化） """
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def bubble_sort_with_flag(nums):
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n = len(nums)
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# 外循环：待排序元素数量为 n-1, n-2, ..., 1
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for i in range(n - 1, -1, -1):
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flag = False # 初始化标志位
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# 内循环：冒泡操作
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for j in range(i):
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if nums[j] > nums[j + 1]:
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# 交换 nums[j] 与 nums[j + 1]
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nums[j], nums[j + 1] = nums[j + 1], nums[j]
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flag = True # 记录交换元素
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if not flag:
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break # 此轮冒泡未交换任何元素，直接跳出
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```
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@ -44,6 +44,22 @@ comments: true
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}
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```
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=== "Python"
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```python
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""" 插入排序 """
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def insertion_sort(nums):
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# 外循环：base = nums[1], nums[2], ..., nums[n-1]
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for i in range(1, len(nums)):
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base = nums[i]
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j = i - 1
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# 内循环：将 base 插入到左边的正确位置
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while j >= 0 and nums[j] > base:
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nums[j + 1] = nums[j] # 1. 将 nums[j] 向右移动一位
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j -= 1
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nums[j + 1] = base # 2. 将 base 赋值到正确位置
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```
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## 算法特性
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**时间复杂度 $O(n^2)$ ：** 最差情况下，各轮插入操作循环 $n - 1$ , $n-2$ , $\cdots$ , $2$ , $1$ 次，求和为 $\frac{(n - 1) n}{2}$ ，使用 $O(n^2)$ 时间。
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@ -7,7 +7,6 @@ comments: true
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「归并排序 Merge Sort」是算法中 “分治思想” 的典型体现，其有「划分」和「合并」两个阶段：
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1. **划分阶段：** 通过递归不断 **将数组从中点位置划分开**，将长数组的排序问题转化为短数组的排序问题；
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2. **合并阶段：** 划分到子数组长度为 1 时，开始向上合并，不断将 **左、右两个短排序数组** 合并为 **一个长排序数组**，直至合并至原数组时完成排序；
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@ -104,6 +103,51 @@ comments: true
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}
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```
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=== "Python"
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```python title="merge_sort.py"
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"""
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合并左子数组和右子数组
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左子数组区间 [left, mid]
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右子数组区间 [mid + 1, right]
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"""
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def merge(nums, left, mid, right):
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# 初始化辅助数组 借助 copy模块
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tmp = nums[left:right + 1]
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# 左子数组的起始索引和结束索引
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left_start, left_end = left - left, mid - left
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# 右子数组的起始索引和结束索引
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right_start, right_end = mid + 1 - left, right - left
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# i, j 分别指向左子数组、右子数组的首元素
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i, j = left_start, right_start
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# 通过覆盖原数组 nums 来合并左子数组和右子数组
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for k in range(left, right + 1):
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# 若 “左子数组已全部合并完”，则选取右子数组元素，并且 j++
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if i > left_end:
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nums[k] = tmp[j]
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j += 1
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# 否则，若 “右子数组已全部合并完” 或 “左子数组元素 < 右子数组元素”，则选取左子数组元素，并且 i++
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elif j > right_end or tmp[i] <= tmp[j]:
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nums[k] = tmp[i]
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i += 1
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# 否则，若 “左子数组元素 > 右子数组元素”，则选取右子数组元素，并且 j++
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else:
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nums[k] = tmp[j]
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j += 1
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""" 归并排序 """
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def merge_sort(nums, left, right):
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# 终止条件
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if left >= right:
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return # 当子数组长度为 1 时终止递归
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# 划分阶段
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mid = (left + right) // 2 # 计算中点
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merge_sort(nums, left, mid) # 递归左子数组
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merge_sort(nums, mid + 1, right) # 递归右子数组
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# 合并阶段
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merge(nums, left, mid, right)
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```
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下面重点解释一下合并方法 `merge()` 的流程：
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1. 初始化一个辅助数组 `tmp` 暂存待合并区间 `[left, right]` 内的元素，后续通过覆盖原数组 `nums` 的元素来实现合并；
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@ -61,6 +61,25 @@ comments: true
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}
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```
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=== "Python"
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```python title="quick_sort.py"
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""" 哨兵划分 """
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def partition(self, nums, left, right):
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# 以 nums[left] 作为基准数
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i, j = left, right
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while i < j:
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while i < j and nums[j] >= nums[left]:
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j -= 1 # 从右向左找首个小于基准数的元素
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while i < j and nums[i] <= nums[left]:
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i += 1 # 从左向右找首个大于基准数的元素
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# 元素交换
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nums[i], nums[j] = nums[j], nums[i]
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# 将基准数交换至两子数组的分界线
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nums[i], nums[left] = nums[left], nums[i]
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return i # 返回基准数的索引
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```
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!!! note "快速排序的分治思想"
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哨兵划分的实质是将 **一个长数组的排序问题** 简化为 **两个短数组的排序问题**。
|
||||
@ -93,6 +112,21 @@ comments: true
|
||||
}
|
||||
```
|
||||
|
||||
=== "Python"
|
||||
|
||||
```python title="quick_sort.py"
|
||||
""" 快速排序 """
|
||||
def quick_sort(self, nums, left, right):
|
||||
# 子数组长度为 1 时终止递归
|
||||
if left >= right:
|
||||
return
|
||||
# 哨兵划分
|
||||
pivot = self.partition(nums, left, right)
|
||||
# 递归左子数组、右子数组
|
||||
self.quick_sort(nums, left, pivot - 1)
|
||||
self.quick_sort(nums, pivot + 1, right)
|
||||
```
|
||||
|
||||
## 算法特性
|
||||
|
||||
**平均时间复杂度 $O(n \log n)$ ：** 平均情况下，哨兵划分的递归层数为 $\log n$ ，每层中的总循环数为 $n$ ，总体使用 $O(n \log n)$ 时间。
|
||||
@ -149,6 +183,29 @@ comments: true
|
||||
}
|
||||
```
|
||||
|
||||
=== "Python"
|
||||
|
||||
```python title="quick_sort.py"
|
||||
""" 选取三个元素的中位数 """
|
||||
def median_three(self, nums, left, mid, right):
|
||||
# 使用了异或操作来简化代码
|
||||
# 异或规则为 0 ^ 0 = 1 ^ 1 = 0, 0 ^ 1 = 1 ^ 0 = 1
|
||||
if (nums[left] > nums[mid]) ^ (nums[left] > nums[right]):
|
||||
return left
|
||||
elif (nums[mid] < nums[left]) ^ (nums[mid] > nums[right]):
|
||||
return mid
|
||||
return right
|
||||
|
||||
""" 哨兵划分（三数取中值） """
|
||||
def partition(self, nums, left, right):
|
||||
# 以 nums[left] 作为基准数
|
||||
med = self.median_three(nums, left, (left + right) // 2, right)
|
||||
# 将中位数交换至数组最左端
|
||||
nums[left], nums[med] = nums[med], nums[left]
|
||||
# 以 nums[left] 作为基准数
|
||||
# 下同省略...
|
||||
```
|
||||
|
||||
## 尾递归优化
|
||||
|
||||
**普通快速排序在某些输入下的空间效率变差。** 仍然以完全倒序的输入数组为例，由于每轮哨兵划分后右子数组长度为 0 ，那么将形成一个高度为 $n - 1$ 的递归树，此时使用的栈帧空间大小劣化至 $O(n)$ 。
|
||||
@ -175,3 +232,21 @@ comments: true
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
=== "Python"
|
||||
|
||||
```python title="quick_sort.py"
|
||||
""" 快速排序（尾递归优化） """
|
||||
def quick_sort(self, nums, left, right):
|
||||
# 子数组长度为 1 时终止
|
||||
while left < right:
|
||||
# 哨兵划分操作
|
||||
pivot = self.partition(nums, left, right)
|
||||
# 对两个子数组中较短的那个执行快排
|
||||
if pivot - left < right - pivot:
|
||||
self.quick_sort(nums, left, pivot - 1) # 递归排序左子数组
|
||||
left = pivot + 1 # 剩余待排序区间为 [pivot + 1, right]
|
||||
else:
|
||||
self.quick_sort(nums, pivot + 1, right) # 递归排序右子数组
|
||||
right = pivot - 1 # 剩余待排序区间为 [left, pivot - 1]
|
||||
```
|
||||
|
||||
Loading…
Reference in New Issue
Block a user