Leetcode 200 岛屿的数量：

DFS利用函数调用栈保证了检索顺序，

BFS则需要自己建立队列，把待检索对象按规则入队。

class Solution {  // DFS解法，8ms/10.7MB，99.7% / 92%public:    /**    * row,col: 坐标，以0开始.    * 用row和col，而不是x，y. 否则容易写成grid[x][y],顺序不对！！    */    void dfs_set_zero(vector
    
     
      >& grid, int row, int col) {        int n_rows = grid.size();        if (!n_rows) return;        int n_cols = grid[0].size();        if (!n_cols) return;                if (row<0 || row>=n_rows || col<0 || col>=n_cols) return;                if (grid[row][col] == '1') {            grid[row][col] = '0';            dfs_set_zero(grid, row-1, col);            dfs_set_zero(grid, row+1, col);            dfs_set_zero(grid, row, col-1);            dfs_set_zero(grid, row, col+1);        }    }    int numIslands(vector
      
       
        >& grid) {        int result = 0;        for (int i=0; i
       
      
     
    

class Solution {  // BFS解法，12ms/11MB, 96.9% / 69.4%public:    queue
    
     
      > q;        void bfs_set_zero(vector
      
       
        >& grid) {        int n_rows = grid.size();        if(!n_rows) return;        int n_cols = grid[0].size();        if(!n_cols) return;                while (!q.empty()) {            auto pos = q.front();            q.pop();            //if (grid[pos.first][pos.second] == '1') {                grid[pos.first][pos.second] = '0';                if (pos.first-1>=0 && grid[pos.first-1][pos.second] == '1')                    q.emplace(pos.first-1,pos.second);                if (pos.first+1
        
         =0 && grid[pos.first][pos.second-1] == '1')                    q.emplace(pos.first,pos.second-1);                if (pos.second+1
         
          >& grid) { int result = 0; for (int i=0; i