思路：利用深度优先搜索的思路来查找1身边的1，并且遍历之后进行0替换防止重复dfs，代码如下所示

class Solution {
public:int numIslands(vector<vector<char>>& grid) {int row = grid.size();int col = grid[0].size();int numoflands = 0;//从第一行开始遍历，是岛屿则进行周围深度优先搜索for(int r = 0;r < row;r++){for(int c = 0;c < col;c++){if(grid[r][c] == '1'){++numoflands;dfs(grid,r,c);}}}return numoflands;}//dfs用于消除“1”旁边的“1”并用“0”替换掉防止二次遍历void dfs(vector<vector<char>>& grid,int r,int c){int nr = grid.size();int nc = grid[0].size();grid[r][c] = '0';//用“0”替换“1”，防止重复调用if(r-1>=0&&grid[r-1][c] == '1') dfs(grid,r-1,c);if(r+1<nr&&grid[r+1][c] == '1') dfs(grid,r+1,c);if(c-1>=0&&grid[r][c-1] == '1') dfs(grid,r,c-1);if(c+1<nc&&grid[r][c+1] == '1') dfs(grid,r,c+1);}
};