leetcode200. 岛屿数量（深搜dfs）

链接：https://leetcode-cn.com/problems/number-of-islands/

题目

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

用例

示例 1：

输入：grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出：1
示例 2：

输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

思路

遍历地图 检索到为1的格子 岛屿数num++
然后从该格子开始dfs感染周边联通的格子改为'2' 递归
继续遍历

  class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int sum=0;
        int n=grid.size(),m=grid[0].size();
        for(int i=0;i<n;++i)
        {
            for(int j=0;j<m;++j){
                if(grid[i][j]=='1')
                {
                    sum++;
                    infect(i,j,grid,n,m);
                }
            }
        }
        return sum;
    }
    void infect(int i,int j,vector<vector<char>>&grid,int n,int m)
    {
        if(i<0||i>n-1||j<0||j>m-1)
            return;
        if(grid[i][j]!='1')
            return;
        else
            grid[i][j]='2';
        infect(i+1,j,grid,n,m);
        infect(i,j+1,grid,n,m);
        infect(i,j-1,grid,n,m);
        infect(i-1,j,grid,n,m);
    }
};