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：
代码语言：javascript复制输入：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：
代码语言：javascript复制输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3思路 讲述看到这一题的思路
解题方法 描述你的解题方法
复杂度 时间复杂度: 添加时间复杂度, 示例：
空间复杂度: 添加空间复杂度, 示例：
class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
int ans=0,Nh=grid.size(),Nl=grid[0].size();
//可递归的lamber表达式写法
auto dfs=[&](auto & dfs,int r,int c) ->void
{
grid[r][c]='0';
if (r - 1 >= 0 && grid[r-1][c] == '1') dfs(dfs, r-1, c);
if (r 1 < Nh && grid[r 1][c] == '1') dfs(dfs, r 1, c);
if (c - 1 >= 0 && grid[r][c-1] == '1') dfs(dfs, r, c-1);
if (c 1 < Nl && grid[r][c 1] == '1') dfs(dfs, r, c 1);
};
//程序实现。一个个搜索1，并且把一的岛屿清零，然后统计岛屿
for(int i=0;i<Nh;i )
for(int j=0;j<Nl;j )
if(grid[i][j] == '1')
ans ,dfs(dfs,i,j);
return ans;
}
};
