1. Fix the import error.
2. Some codes fine tuning.
This commit is contained in:
@@ -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**。
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@@ -93,6 +112,21 @@ 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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""" 快速排序 """
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def quick_sort(self, nums, left, right):
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# 子数组长度为 1 时终止递归
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if left >= right:
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return
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# 哨兵划分
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pivot = self.partition(nums, left, right)
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# 递归左子数组、右子数组
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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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```
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## 算法特性
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**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。
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@@ -149,6 +183,29 @@ 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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""" 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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# 异或规则为 0 ^ 0 = 1 ^ 1 = 0, 0 ^ 1 = 1 ^ 0 = 1
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if (nums[left] > nums[mid]) ^ (nums[left] > nums[right]):
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return left
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elif (nums[mid] < nums[left]) ^ (nums[mid] > nums[right]):
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return mid
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return right
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""" 哨兵划分(三数取中值) """
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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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med = self.median_three(nums, left, (left + right) // 2, right)
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# 将中位数交换至数组最左端
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nums[left], nums[med] = nums[med], nums[left]
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# 以 nums[left] 作为基准数
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# 下同省略...
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```
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## 尾递归优化
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**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 $O(n)$ 。
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@@ -175,3 +232,21 @@ comments: true
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}
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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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""" 快速排序(尾递归优化) """
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def quick_sort(self, nums, left, right):
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# 子数组长度为 1 时终止
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while left < right:
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# 哨兵划分操作
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pivot = self.partition(nums, left, right)
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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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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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```
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