215. Kth Largest Element in an Array

题意：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.

Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.

思路：这题的方法有很多，特别是O(n^2)的方法，只要是排序的方法都可以解，求出排序的结果，再求-k位索引的值就可以了，但是如果要用O(n)的方法，可能就需要借助快排了，快排每次能将一个放在正确的位置上，快排的具体算法步骤可以参考这个视频舞动的快排 ，很形象的表现了快排的过程，先放一个快排的代码：

class Solution:    # @param {integer} s    # @param {integer[]} nums    # @return {integer}    def quicksort(self, nums, l, r):        if l >= r:            return        pos = self.partition(nums, l, r)        self.quicksort(nums, l, pos-1)        self.quicksort(nums, pos+1, r)    # choose the right-most element as pivot    def partition(self, nums, l, r):        low = l        while l < r:            if nums[l] < nums[r]:                nums[l], nums[low] = nums[low], nums[l]                low += 1            l += 1        nums[low], nums[r] = nums[r], nums[low]        return lowif __name__ == "__main__":    a = [3,4,2,4,5,6,8,3,56,7,8,9,4,3,2]    Solution().quicksort(a,0,len(a)-1)    print a

了解快排之后，就可以来做题了，具体思路是先找到数组末端的数，将它放到正确的位置pos，再到剩下的部分去搜索第k-pos-1或者第k大的结果即可。

def findKthLargest(self, nums, k):    # convert the kth largest to smallest    return self.findKthSmallest(nums, len(nums)+1-k)def findKthSmallest(self, nums, k):    if nums:        pos = self.partition(nums, 0, len(nums)-1)        if k > pos+1:            return self.findKthSmallest(nums[pos+1:], k-pos-1)        elif k < pos+1:            return self.findKthSmallest(nums[:pos], k)        else:            return nums[pos]# choose the right-most element as pivot   def partition(self, nums, l, r):    low = l    while l < r:        if nums[l] < nums[r]:            nums[l], nums[low] = nums[low], nums[l]            low += 1        l += 1    nums[low], nums[r] = nums[r], nums[low]    return low