python实现leetcode 数组中第k个最大的值（快速排序）

# -*- coding: UTF-8 -*-
lst = []
temp = input()

def swap(num,i,j):
    k = num[i]
    num[i] = num[j]
    num[j] = k
    return num

def partition(nums , left,right):
    if left>right:
        return 0
    pivotkey = nums[left]
    while(left<right):
        while(left<right and nums[right] <= pivotkey):
            right -=1
        #swap(nums[left],nums[right])
        nums = swap(nums,left,right)
        print('num=',nums)
        while(left<right and nums[left] >= pivotkey):
            left +=1
        nums = swap(nums,left,right)
    return left

def find(nums,k):
    n = len(nums)
    if (k<1 or k >n):
        return -1
    left = 0
    right = n-1
    while True:
        print('left=',left)
        pos = partition(nums,left,right)
        if pos+1 == k :return nums[pos]
        elif (pos+1 >k):right = pos-1
        else: left = pos+1

if temp != '':
    lst.append([int(i) for i in temp.split(' ')])
    lst1 = lst[0]
    print('lst=',lst1)
    k = input()
    print('k=',k)
    find_num = find(lst1,int(k))
    print('find_num=',find_num)
else:
    print('There is no list')

python版本3.5

结果如图：

 键盘输入4 3 5 2 6 1，k=2，寻找第二大的数，find_name为最终输出结果。