leetcode_378. Kth Smallest Element in a Sorted Matrix_堆的应用

Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.

Note that it is the kth smallest element in the sorted order, not the kth distinct element.

Example:

matrix = [
   [ 1,  5,  9],
   [10, 11, 13],
   [12, 13, 15]
],
k = 8,

return 13.

Note: 
You may assume k is always valid, 1 ≤ k ≤ n2.

矩阵元素按行按列递增有序，将matr[0][0]元素压进堆中，然后进行k-1次下面操作：

　　取堆顶元素，将其在矩阵中的右边和下边的元素入堆，pop堆顶元素。

最后堆顶元素即是第k小的数。

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;

struct Node
{
    int x,y,num;
    Node(int a,int b,int c)
    {
        x=a;
        y=b;
        num=c;
    }
    bool operator < (const Node& x)const
    {
        return num>x.num;
    }
};

class Solution
{
public:
    int kthSmallest(vector<vector<int> >& matrix, int k)
    {
        int len=matrix.size();
        bool inheap[1000][1000];
        memset(inheap,0,sizeof(inheap));
        priority_queue<Node> heap;
        heap.push(Node(0,0,matrix[0][0]));
        for(int i=0; i<k-1; i++)
        {
            Node now=heap.top();
            if(now.x<len-1&&inheap[now.x+1][now.y]==0)
            {
                heap.push(Node(now.x+1,now.y,matrix[now.x+1][now.y]));
                inheap[now.x+1][now.y]=1;
            }
            if(now.y<len-1&&inheap[now.x][now.y+1]==0)
            {
                heap.push(Node(now.x,now.y+1,matrix[now.x][now.y+1]));
                inheap[now.x][now.y+1]=1;
            }
            heap.pop();
        }
        return heap.top().num;
    }
};