题解 | #寻找第K大#

# -*- coding:utf-8 -*-

class Solution:
    
    
    def qu(self,nums,s,e):
        if s>e:
            return 
        mid= nums[s]
        low=s
        high=e 
        while(low<high):
            while low<high and nums[high]>mid:
                high-=1
            nums[low]=nums[high]
            while low<high and nums[low]<=mid:
                low+=1
            nums[high]=nums[low]
        nums[low]=mid
        self.qu(nums,s,low-1)
        self.qu(nums,low+1,e)
    def findKth(self, a, n, K):
        # write code here
        self.qu(a,0,n-1)
        return a[::-1][K-1]