【DFS】200. Number of Islands

采用DFS的思想，遍历1的聚类簇个数；

class Solution {
public:
	void dfs(vector<vector<char>>& grid,vector<vector<bool>>& visit, int row, int col)
	{
		int dx[4]={0,-1,0,1};
		int dy[4]={-1,0,1,0};
		visit[row][col]=true;	//开始遍历当前节点,标记为已遍历状态
		for(int k=0;k<4;k++)
		{
			int newRow=row+dx[k];
			int newCol=col+dy[k];
			if(newRow>-1&&newRow<grid.size()&&newCol>-1&&newCol<grid[0].size()&&grid[newRow][newCol]=='1'&&!visit[newRow][newCol])
			{  //遍历条件：下标未越界，节点value为小岛领土且节点未被遍历
				dfs(grid,visit,newRow,newCol);	//遍历相邻四个节点
			}
		}
	}
    int numIslands(vector<vector<char>>& grid)
    {
        if(grid.size()==0)
            return 0;
    	int ans=0;
    	vector<vector<bool>> visit(grid.size(),vector<bool>(grid[0].size(),false));
    	for(int i=0;i<grid.size();i++)
    	{
    		for(int j=0;j<grid[0].size();j++)
    		{
    			if(visit[i][j]||grid[i][j]=='0')
    				continue;
    			ans++;  //聚类簇个数增一 
    			dfs(grid,visit,i,j);  //当前节点未遍历且值为1的情况下，开始进行遍历
    		}
    	}
    	return ans;
    }
};