Number of Islands —— Leetcode（重要的一类题）

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

深度优先的递归算法比广度优先快那么多，应该是queue造成的，代码如下：

class Solution {
public:
    void BFS(vector<vector<char>>& grid, int x, int y) {
        queue<vector<int>> q;
        q.push({x, y});
        grid[x][y] = '0';
        
        while(!q.empty()) {
            x = q.front()[0];
            y = q.front()[1];
            q.pop();
            
            if(x>0 && grid[x-1][y]=='1') {
                grid[x-1][y] = '0';
                q.push({x-1, y});
            }
            if(x<grid.size()-1 && grid[x+1][y]=='1') {///
                grid[x+1][y] = '0';
                q.push({x+1, y}); 
            } 
            if(y>0 && grid[x][y-1]=='1') {
                grid[x][y-1] = '0';
                q.push({x, y-1}); 
            } 
            if(y<grid[0].size()-1 && grid[x][y+1]=='1') {
                grid[x][y+1] = '0';
                q.push({x, y+1}); 
            }
            
        }
    }
    
    void DFS(vector<vector<char>>& grid, int x, int y) {
        grid[x][y] = '0';
        
        if(x>0 && grid[x-1][y] == '1')
            DFS(grid, x-1, y);
        if(x<grid.size()-1 && grid[x+1][y] == '1')
            DFS(grid, x+1, y);
        if(y>0 && grid[x][y-1] == '1')
            DFS(grid, x, y-1);
        if(y<grid[0].size()-1 && grid[x][y+1] == '1')
            DFS(grid, x, y+1);
    }
    
    int numIslands(vector<vector<char>>& grid) {
        if(grid.size()==0 || grid[0].size()==0)
            return 0;
            
        int res = 0;
        for(int i=0; i<grid.size(); i++) {
            for(int j=0; j<grid[0].size(); j++) {
                if(grid[i][j] == '1') {
                    ++ res;
                    DFS(grid, i, j);    //击败100%的C++ coder
                    //BFS(grid, i, j);  //击败12.56%的C++ coder
                }
            }
        }
        return res;
    }
};