队列 & 栈//岛屿的个数

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 numIslands(char[][] grid) {
        if(grid.length==0||grid[0].length==0) return 0;
        int m = grid.length;
        int n = grid[0].length;
        int res = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == '1')
                {
                    dfs(grid,i,j);
                    res++;
                }
            }
        }
        return res;
    }
    public void dfs(char[][] g, int x, int y){
        if(x<0||x>=g.length) return ;
        if(y<0||y>=g[0].length) return ;
        if(g[x][y]!='1')
            return ;
        g[x][y] = '0';
        dfs(g,x+1,y);
        dfs(g,x-1,y);
        dfs(g,x,y+1);
        dfs(g,x,y-1);
    }
}

class Solution {
public:
    int numIslands(vector<vector<char> > &grid) {
        if (grid.empty() || grid[0].empty()) return 0;
        int m = grid.size(), n = grid[0].size(), res = 0;
        vector<vector<bool> > visited(m, vector<bool>(n, false));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1' && !visited[i][j]) {
                    numIslandsDFS(grid, visited, i, j);
                    ++res;
                }
            }
        }
        return res;
    }
    void numIslandsDFS(vector<vector<char> > &grid, vector<vector<bool> > &visited, int x, int y) {
        if (x < 0 || x >= grid.size()) return;
        if (y < 0 || y >= grid[0].size()) return;
        if (grid[x][y] != '1' || visited[x][y]) return;
        visited[x][y] = true;
        numIslandsDFS(grid, visited, x - 1, y);
        numIslandsDFS(grid, visited, x + 1, y);
        numIslandsDFS(grid, visited, x, y - 1);
        numIslandsDFS(grid, visited, x, y + 1);
    }
};