Kth Largest Element in an Array

题目描述：

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.

题目大意：

在未排序的数组中找出第k大的元素。（k一定是有效的， 1 ≤ k ≤ 数组长度）

注意是排序之后的第k大元素，不是第k个不重复的元素

示例：

给定数组[3,2,1,5,6,4]，k = 2，return 5.

解题思路：

（1）堆排序(O(nlogn))，使用数组的前k个元素创建一个最小堆min_heap，剩下的的n-k个元素依次插入堆，并弹出最小值，保持堆大小为k，最后取min_heap[0]即为第k大的元素。

class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        from heapq import heappush, heappushpop
        heap = []
        for i in xrange(k):
            heappush(heap, nums[i])
            
        for i in xrange(k, len(nums)):
            heappushpop(heap, nums[i])
            
        return heap[0]

（2）快速选择排序(O(n))，参考 http://www.cs.yale.edu/homes/aspnes/pinewiki/QuickSelect.html

from random import choice

class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        # 随机选取一个数字
        pivot = choice(nums)
        # 比pivot分别大的小的放到不同的a1,a2
        a1, a2 = [], []
        for i in nums:
            if i > pivot:
                a1.append(i)
            elif i < pivot:
                a2.append(i)
        
        # a1的长度大于k，递归a1      
        if k <= len(a1):
            return self.findKthLargest(a1, k)
        # a2的长度大于len(nums) - k，说明第k大元素为a2的第k-(len(nums)-len(a2))大元素
        elif k > len(nums) - len(a2):
            return self.findKthLargest(a2, k - (len(nums) - len(a2)))
        
        # k = len(nums) - len(a2), pivot为第k大的元素
        return pivot