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;

     }

 };