动态git可以看 :https://leetcode.cn/problems/sort-an-array/solutions/179370/python-shi-xian-de-shi-da-jing-dian-pai-xu-suan-fa/

选择排序

def selection_sort(nums):
    n = len(nums)
    for i in range(n):
        for j in range(i, n):
            if nums[i] > nums[j]:
                nums[i], nums[j] = nums[j], nums[i]
    return nums

为什么只交换严格小于的呢?

插入排序

class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        for i in range(1, len(nums)):
            while i > 0 and nums[i] < nums[i - 1]:
                nums[i-1], nums[i] = nums[i], nums[i-1]
                i -= 1
        return nums

注意使用里面的while loop来实现一直交换的效果

冒泡排序

class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        for i in range(len(nums)):
            swapped = False
            for j in range(1, len(nums) - i):
                if nums[j] < nums[j - 1]:
                    nums[j], nums[j - 1] = nums[j - 1], nums[j]
                    swapped = True
            if not swapped:
                break
        return nums

注意两个点: 1.用最外层的for loop代表最后n个数字已经完成了排序 2. 用swap变量来提前终止循环

merge sort

def merge_sort(nums):
    if len(nums) <= 1:
        return nums
    mid = len(nums) // 2
    # 分
    left = merge_sort(nums[:mid])
    right = merge_sort(nums[mid:])
    # 合并
    return merge(left, right)


def merge(left, right):
    res = []
    i = 0
    j = 0
    while i < len(left) and j < len(right):
        if left[i] <= right[j]:
            res.append(left[i])
            i += 1
        else:
            res.append(right[j])
            j += 1
    res += left[i:]
    res += right[j:]
    return res

注意不要忘记了base case哦

快速排序 (注意下partition那一步是怎么写的)

def quick_sort(nums):
    n = len(nums)

    def quick(left, right):
        if left >= right:
            return nums
        pivot = left
        i = left
        j = right
        while i < j:
            while i < j and nums[j] >= nums[pivot]:
                j -= 1
            while i < j and nums[i] <= nums[pivot]:
                i += 1
            nums[i], nums[j] = nums[j], nums[i]
        nums[pivot], nums[j] = nums[j], nums[pivot]
        quick(left, j - 1)
        quick(j + 1, right)
        return nums

    return quick(0, n - 1)

一定要注意是先变化j, 因为我们的pivot是从index 0 选的, 就应该从对面开始

215 数组中第k大的元素

就是quick_select