LeetCode 542 01 Matrix BFS

Given an m x n binary matrix mat, return the distance of the nearest 0 for each cell.

The distance between two adjacent cells is \(1\)

Solution

首先将所有的 \(dis\) 令成 \(-1\)，然后把所有 \(0\) 的点 \(push\) 到队列里面，每次更新 \(dis[i][j]=-1\) 的点

点击查看代码

class Solution {
private:
    int dir[4][2] = {0,1, 1,0, 0,-1, -1,0};
    bool check(int x,int y,int r, int c){
        if(x<0||y<0||x>=r||y>=c)return false;
        return true;
    }
    
    
    queue<vector<int>> q;
    
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int r = mat.size(), c = mat[0].size();
        vector<vector<int>> dis(r, vector<int>(c,-1));
        for(int i=0;i<r;i++){
            for(int j=0;j<c;j++){
                if(mat[i][j]==0){
                    dis[i][j]=0;q.push({i,j});
                }
            }
        }
        
        while(!q.empty()){
            auto f = q.front();q.pop();
            int x = f[0], y = f[1];
            
            for(int i=0;i<4;i++){
                int nx = x+dir[i][0], ny = y+dir[i][1];
                if(check(nx,ny,r,c)&&dis[nx][ny]==-1){
                    dis[nx][ny] = dis[x][y]+1;
                    q.push({nx,ny});
                }
            }
        }
        return dis;
    }
};