DFS或BFS(深度优先搜索或广度优先搜索遍历无向图)-04-无向图-岛屿数量

给定一个由 '1'（陆地）和 '0'（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。

示例 1:

输入:
11110
11010
11000
00000

输出: 1
示例 2:

输入:
11000
11000
00100
00011

输出: 3

class Solution {
public:
    int dist[4][2] = {0,1,0,-1,1,0,-1,0};
    queue<pair<int,int>>que;
    void dfs(vector<vector<char>>& grid,vector<vector<int>>& sign,int i,int j,int row,int col,int num){
        sign[i][j] = 1;
        int ii,jj;
        for(int zz = 0;zz < 4;zz++){
            ii = i + dist[zz][0];
            jj = j + dist[zz][1];
            if(ii < 0||jj < 0||ii >= row||jj >= col)  continue;
            if(sign[ii][jj] == 1) continue;
            if(grid[ii][jj] == '1')
                dfs(grid,sign,ii,jj,row,col,num);
        }
    }
    void bfs(vector<vector<char>>& grid,vector<vector<int>>& sign,int i,int j,int row,int col,int num){
        sign[i][j] = 1;
        int ii,jj;
        for(int zz = 0;zz < 4;zz++){
            ii = i + dist[zz][0];
            jj = j + dist[zz][1];
            if(ii < 0||jj < 0||ii >= row||jj >= col)  continue;
            if(sign[ii][jj] == 1) continue;
            if(grid[ii][jj] == '1'){
                sign[ii][jj] = 1;
                que.push({ii,jj});
            }
        }
        while(!que.empty()){
            ii = que.front().first;
            jj = que.front().second;
            que.pop();
            bfs(grid,sign,ii,jj,row,col,num);
        }
        
    }
    int numIslands(vector<vector<char>>& grid) {
        if(grid.size() == 0||grid[0].size() == 0)
            return 0;
        int row = grid.size();
        int col = grid[0].size();
        int num = 0;
        vector<vector<int>>sign(row,vector<int>(col,0));
        for(int i = 0 ;i < row;i++){
            for(int j = 0;j < col;j++){
                if(sign[i][j] == 0 && grid[i][j] == '1'){
                    num++;
                    dfs(grid,sign,i,j,row,col,num);
                }
            }
        }
        return num;
    }
};