python快速排序及其应用_python快速排序用处-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_37591364/article/details/89163826

本文探讨了Python的快速排序算法，并通过实例介绍了如何利用快速排序解决实际问题，包括0-1串的最少交换次数、数组元素奇偶排序以及寻找数组中第k大的数等应用场景。

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1、快速排序

def quicksort(arr):
    length = len(arr)
    if length < 2:
        return arr
    midarr = arr[int(length/2)]
    left = [x for x in arr if x < midarr]
    middle = [x for x in arr if x == midarr]
    right = [x for x in arr if x > midarr]
    return quicksort(left) +middle +quicksort(right)

def partition(A, left,right):
    tmp = A[left]
    while left < right:
        while left < right and A[right] >= tmp:
            right -= 1
        A[left] = A[right]
        while left < right and A[left] <= tmp:
            left += 1
        A[right] = A[left]
    A[left] = tmp
    return left
def quick_sort(A,left, right):
    if left < right:
        pos = partition(A, left, right)
        quick_sort(A,left, pos-1)
        quick_sort(A,pos+1, right)
    return A

2、把一个0-1串（只包含0和1的串）进行排序，你可以交换任意两个位置，问最少交换的次数？

def changestr(str):
    arr = list(str)
    i = 0
    j = len(arr) - 1
    ret = 0
    while i < j:
        # if i == j:
        #     return ''.join(arr)
        while arr[i] == '0':
            i += 1
        while arr[j] == '1':
            j -= 1
        if i < j:
            tmp = arr[i]
            arr[i] =arr[j]
            arr[j] = tmp
            ret += 1
    return ''.join(arr),ret

3、实现对数组奇数在前偶数在后

def changesort(arr):
    length = len(arr)
    i = 0
    j = length - 1
    while i < j:
        while arr[i] % 2 == 1:
            i += 1
        while arr[j] % 2 == 0:
            j -= 1
        if i < j:
            tmp = arr[i]
            arr[i] = arr[j]
            arr[j] = tmp
    return arr
print(changesort([1,8,5,3,4,2,9,7,5]))

4、寻找数组中第k大的数

def partition(A, left,right):
    tmp = A[left]
    while left < right:
        while left < right and A[right] >= tmp:
            right -= 1
        A[left] = A[right]
        while left < right and A[left] <= tmp:
            left += 1
        A[right] = A[left]
    A[left] = tmp
    return left

def findkth(A,left,right,k):
    pos = partition(A, left,right)
    if pos == k:
        return A[pos]
    elif pos < k:
        return findkth(A,pos+1,right,k)
    else:
        return findkth(A, left, pos-1, k)