(BFS/DFS) leetcode 200. Number of Islands

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:

Input:
11110
11010
11000
00000

Output: 1

Example 2:

Input:
11000
11000
00100
00011

Output: 3

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顾名思义，就是求岛屿的个数，此题可以用bfs，也可以用dfs求解，相对来说，用dfs写代码会更简单，但是这个题我用bfs。

这个题用bfs时，不必新设立vis来判断是否访问过，可以把访问到的‘1’的字符中，把这个字符改为'0'就可以。

 

C++代码：

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        typedef pair<int,int> pii;
        queue<pii> q;
        int x = ,y = ,xx = ,yy = ;
        int dx[] = {,-,,};
        int dy[] = {,,,-};
        int row = grid.size();
        int col;
        int sum = ;
        if(row > ){
            col = grid[].size();
        }
        else{
            col = ;
        }
        if(row ==  || col == ){
            return ;
        }
        for(int i = ; i < row; i++){
            for(int j = ; j < col; j++){
                if(grid[i][j] == ''){
                    grid[i][j] == '';
                    q.push(pii(i,j));
                    while(!q.empty()){
                        x = q.front().first;
                        y = q.front().second;
                        q.pop();
                        for(int i = ; i < ; i++){
                            xx = x + dx[i];
                            yy = y + dy[i];
                            if(xx >=  && xx < row && yy >=  && yy < col && grid[xx][yy] == ''){
                                grid[xx][yy] = '';
                                q.push(pii(xx,yy));
                            }
                        }
                    }
                    sum++;
                }
            }
        }
        return sum;
    }
};

 emmmm，虽然用DFS写代码会更简单些，但是。。。我怎么觉得DFS比BFS更难理解

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