示例代码

class Solution {
public:
    queue<pair<int,int>> points;
    vector<int> direc={-1,0,1,0,-1};
    int m,n;
    void dfs(vector<vector<int>>& grid,int r,int c){
        if(grid[r][c]==0){
            points.push({r,c});
            return;
        }
        grid[r][c]=2;
        for(int i=0;i<4;i++){
            int x=r+direc[i],y=c+direc[i+1];
            if(x>=0&&x<m&&y>=0&&y<n&&grid[x][y]!=2){
                dfs(grid,x,y);
            }
        }


    }
    int shortestBridge(vector<vector<int>>& grid) {
        m=grid.size(),n=grid[0].size();
        //dfs寻找第一个岛屿，并把1全部赋值为22
        bool flag=false;
        for(int i=0;i<m;i++){
            if(flag){
                break;
            }
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    dfs(grid,i,j);
                    flag=true;
                    break;
                }
            }
        }

        //bfs寻找第二个岛屿,并把经过的0赋值为2
        int level=0;
        while(!points.empty()){
            level++;
            int len=points.size();
            while(len--){
                auto [r,c]=points.front();
                points.pop();
                for(int i=0;i<4;i++){
                    int x=r+direc[i],y=c+direc[i+1];
                    if(x>=0&&x<m&&y>=0&&y<n&&grid[x][y]!=2){
                        if(grid[x][y]==1){
                            return level;
                        }else{
                            points.push({x,y});
                            grid[x][y]=2;
                        }
                    }
                }
            }
        }

        return 0;

    }
};

效果展示

LeetCode---934.最短的桥（dfs+bfs）_leetcode