Kth Largest Element in an Array

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：方法一：排序，然后输出第k个数。O(nlogn)。

方法二：进行k次冒泡排序，然后输出倒数第k个数。O(nk)。

方法三：构建一个大小为n的大顶堆，然后从中抽取k-1次最大值，之后堆顶的值就是第k大的数。复杂度为O(n + klogn)。

下面附上方法三的代码。

class Solution {
public:
    int heap_size;
    void Max_Heapify(vector<int>& nums, int i)
    {
        int l = i * 2 + 1;
        int r = i * 2 + 2;
        int largest = i;
        if (l < heap_size && nums[l] > nums[i])
            largest = l;
        if (r < heap_size && nums[r] > nums[largest])
            largest = r;
        if (largest != i)
        {
            swap(nums[i], nums[largest]);
            Max_Heapify(nums, largest);
        }
    }
    void Build_MaxHeap(vector<int>& nums, int length)
    {
        heap_size = length;
        for (int i = heap_size / 2 - 1; i >= 0; i--)
            Max_Heapify(nums, i);
    }
    void Extract(vector<int>& nums)
    {
        if (heap_size == 0) return;
        swap(nums[0], nums[heap_size - 1]);
        heap_size--;
        Max_Heapify(nums, 0);
    }
    int findKthLargest(vector<int>& nums, int k) {
        Build_MaxHeap(nums, nums.size());
        for (int i = 1; i < k; i++)
            Extract(nums);
        return nums[0];
    }
};