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快速选择

R
P
"""
A Python implementation of the quick select algorithm, which is efficient for
calculating the value that would appear in the index of a list if it would be
sorted, even if it is not already sorted
https://en.wikipedia.org/wiki/Quickselect
"""

import random


def _partition(data: list, pivot) -> tuple:
    """
    Three way partition the data into smaller, equal and greater lists,
    in relationship to the pivot
    :param data: The data to be sorted (a list)
    :param pivot: The value to partition the data on
    :return: Three list: smaller, equal and greater
    """
    less, equal, greater = [], [], []
    for element in data:
        if element < pivot:
            less.append(element)
        elif element > pivot:
            greater.append(element)
        else:
            equal.append(element)
    return less, equal, greater


def quick_select(items: list, index: int):
    """
    >>> quick_select([2, 4, 5, 7, 899, 54, 32], 5)
    54
    >>> quick_select([2, 4, 5, 7, 899, 54, 32], 1)
    4
    >>> quick_select([5, 4, 3, 2], 2)
    4
    >>> quick_select([3, 5, 7, 10, 2, 12], 3)
    7
    """
    # index = len(items) // 2 when trying to find the median
    #   (value of index when items is sorted)

    # invalid input
    if index >= len(items) or index < 0:
        return None

    pivot = items[random.randint(0, len(items) - 1)]
    count = 0
    smaller, equal, larger = _partition(items, pivot)
    count = len(equal)
    m = len(smaller)

    # index is the pivot
    if m <= index < m + count:
        return pivot
    # must be in smaller
    elif m > index:
        return quick_select(smaller, index)
    # must be in larger
    else:
        return quick_select(larger, index - (m + count))
关于此算法

问题陈述

给定一个数组,在线性时间复杂度内找到第 k 大/小的元素。

方法

  • 随机选择一个枢纽元素
  • 应用快速排序中使用的分区
  • 分区后,枢纽将被放置在其排序位置,即所有小于枢纽的元素将在其左侧,大于枢纽的元素将在其右侧
  • 如果排序后的枢纽索引为 k,则枢纽为我们的第 k 个元素,我们返回该数字
  • 否则,检查“k”是大于还是小于,并在该范围内选择一个新的枢纽。
  • 重复直到我们在第 k 个位置获得第 k 个元素

时间复杂度

  • O(n^2) 最坏情况性能

  • O(n) 最佳情况性能

  • O(n) 平均性能

创始人姓名

  • 该算法由托尼·霍尔开发,也称为霍尔选择算法

示例

arr[] = {8,2,11,7,9,1,3}
Indexes: 0 1 2 3 4 5 6

Let's say k = 4. ie. We have to find 4th smallest element.

1. Choosing random pivot as 7
2. Swap 7 with the last element and apply the partitioning algorithm
3. After applying partition, all elements smaller than 7 will be placed to the left and greater in its right.
   Thus we can say that 7 is in its sorted position arr[] = {2,3,1,7,8,9,11}
4. As position of '7' is 4th (ie. k). Thus we will simply return 7

有用的视频链接

视频解释如何在不同的复杂度下找到第 K 个最小/最大元素