題目

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

思路

利用快速排序的思想，用數(shù)組頭將數(shù)組兩分，左邊比它小，右邊比它大。如果這個(gè)元素正好是第k個(gè)，直接返回；如果比k大，則從比它小的那些數(shù)中找出第k大的；否則，從比它大的數(shù)中找出相應(yīng)的數(shù)字。

代碼

class Solution(object):

    def quick_sort(self, nums, k):
        i = 0
        for j in range(1,len(nums)):
            if nums[j] > nums[0]:
                i += 1
                nums[i],nums[j] = nums[j],nums[i]
        nums[i],nums[0] = nums[0],nums[i]
        if i == k-1:
            return nums[i]
        elif i > k-1:
            return self.quick_sort(nums[:i], k)
        else:
            return self.quick_sort(nums[i+1:], k-i-1)

    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        if nums==None or len(nums)==0 or k<1:
            return -1
        return self.quick_sort(nums, k)