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[][] = {,,,-,,,-,};
     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] = ;
         int ii,jj;
         for(int zz = ;zz < ;zz++){
             ii = i + dist[zz][];
             jj = j + dist[zz][];
             if(ii < ||jj < ||ii >= row||jj >= col)  continue;
             if(sign[ii][jj] == ) continue;
             if(grid[ii][jj] == '')
                 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] = ;
         int ii,jj;
         for(int zz = ;zz < ;zz++){
             ii = i + dist[zz][];
             jj = j + dist[zz][];
             if(ii < ||jj < ||ii >= row||jj >= col)  continue;
             if(sign[ii][jj] == ) continue;
             if(grid[ii][jj] == ''){
                 sign[ii][jj] = ;
                 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() == ||grid[].size() == )
             return ;
         int row = grid.size();
         int col = grid[].size();
         int num = ;
         vector<vector<int>>sign(row,vector<int>(col,));
         for(int i =  ;i < row;i++){
             for(int j = ;j < col;j++){
                 if(sign[i][j] ==  && grid[i][j] == ''){
                     num++;
                     dfs(grid,sign,i,j,row,col,num);
                 }
             }
         }
         return num;
     }
 };