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.

Example 1:

Input: [3,2,1,5,6,4] and k = 2
Output: 5
Example 2:

Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4
Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.

class Solution {
public:
    void maxHeapify(vector<int>& a,int i,int heapSize){
        int l = i*2+1,r = i*2 +2,largest = i;
        if(l < heapSize && a[l] > a[largest]){
            largest = l;
        }
        if(r < heapSize && a[r] > a[largest]){
            largest = r;
        }
        if(largest != i){
            swap(a[i],a[largest]);//把当前节点和孩子节点中最大的节点进行交换
            maxHeapify(a,largest,heapSize);//继续调整孩子节点之后的堆
        }
    }
    void buildMaxHeap(vector<int>& a,int heapSize){
        for(int i = heapSize/2;i>=0; --i){//从底层向上开始建立相应的堆
        maxHeapify(a,i,heapSize);
        }
    }
    int findKthLargest(vector<int>& nums, int k) {
        int heapSize = nums.size();
        buildMaxHeap(nums,heapSize);
        for(int i=nums.size()-1;i>=nums.size()-k+1;--i){
            swap(nums[0],nums[i]);
            --heapSize;
            maxHeapify(nums,0,heapSize);//每次都把最大值用最小值进行覆盖，这样操作便可以将堆里的最大值删除
        }
        return nums[0];
    }
};

注意：1.堆排序的写法，开始位置是0 2.注意每次建立完相应的堆后，都把堆中最大的数删除，然后重新调整堆，直到最后找到相应的数