找出数组第k小的元素

解法

1. 首先容易想到的做 $k$ 次冒泡， 复杂度为 $O(nk)$
2. 或者先排序再找第 $k$ 个元素，复杂度为 $O(n\log n)$
3. 建立 $k$ 个元素的大根堆，遍历后续 $n-k$ 个元素，如果比根小就替换根， 复杂度$O(n\log k)$
4. 利用分割技术 (partitioning)，复杂度为 $O(n\log k)$

迭代形式

import random

def kthSmallest(data, k):
    "Find the nth rank ordered element (the least value has rank 0)."
    data = list(data)
    if not 0 <= k < len(data):
        raise ValueError('not enough elements for the given rank')

    while True:
        pivot = random.choice(data)
        pcount = 0
        low, high = [], []
        for elem in data:
            if elem < pivot:
                low.append(elem)
            elif elem > pivot:
                high.append(elem)
            else:
                pcount += 1
        if k < len(low):
            data = low
        elif k < len(low) + pcount:
            return pivot
        else:
            data = high
            k -= len(low) + pcount

递归形式

def kthSmallest(data, k):

    if not 0 <= k < len(data):
        raise ValueError('not enough elements for the given rank')

    pivot = random.choice(data)
    pcount = 0
    low, high = [], []
    for elem in data:
        if elem < pivot:
            low.append(elem)
        elif elem > pivot:
            high.append(elem)
        else:
            pcount += 1
    if k < len(low):
        return kthSmallest(low, k)
    elif k < len(low) + pcount:
        return pivot
    else:
        return kthSmallest(high, k-len(low)-pcount)

测试

A = [i for i in range(1000)]
random.shuffle(A)
for i in range(len(A)):
    assert(kthSmallest(A, i) == i)

复杂度分析

来源：CSDN

作者：颹蕭蕭

链接：https://blog.csdn.net/itnerd/article/details/103946663