岛屿数量(深度优先遍历)

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

 

class Solution {
public:
    //因为岛屿与岛屿之间是分割的，所以碰到陆地就变为水，便于接下来计算剩余的岛屿，直至没有岛屿
    int m ,n;
    vector<vector<int>> direction={
        {1,0},{-1,0},{0,-1},{0,1}
    };
    void dfs(vector<vector<char>>& grid,int i,int j){
        grid[i][j]='0';//将陆地变为水
        for(int k=0;k<4;k++){
            int x = i+direction[k][0];
            int y = j+direction[k][1];
            if(x>=0&&x<m&&y>=0&&y<n&&grid[x][y]=='1')
                dfs(grid,x,y);
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        m = grid.size();
        n = grid[0].size();
        int res = 0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='1'){
                    dfs(grid,i,j);
                    res++;
                }
            }
        }
        return res;
    }
};