leetcode 200 岛屿数量

深度优先搜索的想法，针对一个无向图，对这一图进行遍历，遇到一个‘1’就进行深度优先搜索，遇到的‘1’都变成‘0’，代表这一个岛屿已经被记录。岛屿的数量就是进行搜索的次数，贴代码

class Solution {
private:
    void dfs(vector<vector<char>>& grid,int x,int y)
    {
        int m = grid.size();
        int n = grid[0].size();
        grid[x][y] = '0';
        if(x-1>=0 && grid[x-1][y] == '1') dfs(grid,x-1,y);
        if(x+1<m && grid[x+1][y] == '1') dfs(grid,x+1,y);
        if(y-1>=0 && grid[x][y-1] == '1') dfs(grid,x,y-1);
        if(y+1<n && grid[x][y+1] == '1') dfs(grid,x,y+1);
    }
public:
    int numIslands(vector<vector<char>>& grid) 
    {
        int m = grid.size();
        if(!m)
        return 0;
        int n = grid[0].size();
        int num = 0;
        for(int i = 0 ; i < m ; i++)
        {
            for(int j = 0 ; j < n ; j++)
            {
                if(grid[i][j] == '1')
                {
                    num++;
                    dfs(grid,i,j);
                }
            }
        }
        return num;
    }
};

广度优先搜索，单次搜索中，通过队列才存储所有周边的‘1’，每搜索到一个‘1’，就将其存入队列中，继续进行搜索，直到队列中元素清空。

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) 
    {
        int m = grid.size();
        int n = grid[0].size();
        int nums = 0;
        for(int i = 0 ; i < m ; i++)
        {
            for(int j = 0 ; j < n ; j++)
            {
                if(grid[i][j] == '1')
                {
                    nums++;
                    grid[i][j] = '0';
                    queue<pair<int,int>> neibor;
                    neibor.push({i,j});
                    while(!neibor.empty())
                    {
                        auto temp = neibor.front();
                        neibor.pop();
                        int x = temp.first;
                        int y = temp.second;
                        if(x-1>=0 && grid[x-1][y] == '1')
                        {
                            neibor.push({x-1,y});
                            grid[x-1][y] = '0';
                        }
                        if(x+1<m && grid[x+1][y] == '1')
                        {
                            neibor.push({x+1,y});
                            grid[x+1][y] = '0';
                        }
                        if(y-1>=0 && grid[x][y-1] == '1')
                        {
                            neibor.push({x,y-1});
                            grid[x][y-1] = '0';
                        }
                        if(y+1<n && grid[x][y+1] == '1')
                        {
                            neibor.push({x,y+1});
                            grid[x][y+1] = '0';
                        }
                    } 
                }
            }
        } 
        return nums; 
    }
};