Python实现快排 𱛡

Posted 2022-08-01 gqy02

原文: http://blog.gqylpy.com/gqy/342

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挖坑法思路：

• 取一个元素p（第一个元素），使元素p归位；
• 列表被p分成两部分，左边的数一定不大于p，右边的数一定不小于p；
• 递归完成排序。

Python代码示例：

lst = [5, 7, 4, 3, 1, 2, 9, 8]


def quick_sort(d, l, r):
    if l < r:
        m = partition(d, l, r)
        quick_sort(d, l, m - 1)
        quick_sort(d, m + 1, r)


def partition(d, l, r):
    # 挖坑法，函数执行结束后，左边的数一定不大于p，右边的数一定不小于p
    p = d[l]
    while l < r:
        while l < r and d[r] >= p:
            r -= 1
        d[l] = d[r]
        while l < r and d[l] <= p:
            l += 1
        d[r] = d[l]
    d[l] = p
    return l


quick_sort(lst, 0, len(lst) - 1)

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《算法导论》中的快速排序：

class QuickSort(object):
    __INSTANCE = None

    def __new__(cls, *args, **kwargs):
        """单例模式"""
        if not cls.__INSTANCE:
            cls.__INSTANCE = object.__new__(cls)
        return cls.__INSTANCE

    def __init__(self, data):
        self.__quick_sort(data, 0, len(data) - 1)

    def __quick_sort(self, d, l, r):
        if l < r:
            q = self.__partition(d, l, r)
            self.__quick_sort(d, l, q - 1)
            self.__quick_sort(d, q + 1, r)

    def __partition(self, d, l, r):
        x = d[r]  # last value
        i = l - 1  # last index
        for j in range(l, r):
            if d[j] <= x:
                i += 1
                d[i], d[j] = d[j], d[i]
        d[i + 1], d[r] = d[r], d[i + 1]
        return i + 1


QuickSort(lst)

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原文: http://blog.gqylpy.com/gqy/342