Kth Largest Element in an Array - LeetCode

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];
    }
};