岛屿数量

题目链接：https://leetcode-cn.com/problems/number-of-islands
题目描述：
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。

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'

题解：

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
       int ans = 0;
       for(int i = 0; i < grid.size(); i++)
       {
           for(int j = 0; j < grid[0].size(); j++)
           {
               if(grid[i][j] == '1')
               {
                   ans++; 
                   dfs(grid, i, j);
               }
           }
       }
       return ans;
    }
    void dfs(vector<vector<char>>& grid, int r, int c)
    {
        //base 判断是否越界
        if(!isLands(grid, r, c))
            return;
        if(grid[r][c] != '1')
            return;
        //标记已经访问过的点
        grid[r][c] = '2';
        //遍历周围点
        dfs(grid, r - 1, c);
        dfs(grid, r + 1, c);
        dfs(grid, r, c - 1);
        dfs(grid, r, c + 1);
    }
    bool isLands(vector<vector<char>>& grid, int r, int c)
    {
        return (r >= 0 && r < grid.size() && c >= 0 && c <grid[0].size());
    }
};