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

 1 class Solution {
 2 public:
 3     int dist[4][2] = {0,1,0,-1,1,0,-1,0};
 4     queue<pair<int,int>>que;
 5     void dfs(vector<vector<char>>& grid,vector<vector<int>>& sign,int i,int j,int row,int col,int num){
 6         sign[i][j] = 1;
 7         int ii,jj;
 8         for(int zz = 0;zz < 4;zz++){
 9             ii = i + dist[zz][0];
10             jj = j + dist[zz][1];
11             if(ii < 0||jj < 0||ii >= row||jj >= col)  continue;
12             if(sign[ii][jj] == 1) continue;
13             if(grid[ii][jj] == '1')
14                 dfs(grid,sign,ii,jj,row,col,num);
15         }
16     }
17     void bfs(vector<vector<char>>& grid,vector<vector<int>>& sign,int i,int j,int row,int col,int num){
18         sign[i][j] = 1;
19         int ii,jj;
20         for(int zz = 0;zz < 4;zz++){
21             ii = i + dist[zz][0];
22             jj = j + dist[zz][1];
23             if(ii < 0||jj < 0||ii >= row||jj >= col)  continue;
24             if(sign[ii][jj] == 1) continue;
25             if(grid[ii][jj] == '1'){
26                 sign[ii][jj] = 1;
27                 que.push({ii,jj});
28             }
29         }
30         while(!que.empty()){
31             ii = que.front().first;
32             jj = que.front().second;
33             que.pop();
34             bfs(grid,sign,ii,jj,row,col,num);
35         }
36         
37     }
38     int numIslands(vector<vector<char>>& grid) {
39         if(grid.size() == 0||grid[0].size() == 0)
40             return 0;
41         int row = grid.size();
42         int col = grid[0].size();
43         int num = 0;
44         vector<vector<int>>sign(row,vector<int>(col,0));
45         for(int i = 0 ;i < row;i++){
46             for(int j = 0;j < col;j++){
47                 if(sign[i][j] == 0 && grid[i][j] == '1'){
48                     num++;
49                     dfs(grid,sign,i,j,row,col,num);
50                 }
51             }
52         }
53         return num;
54     }
55 };