岛屿数量(力扣200)-DFS

题目:岛屿数量

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。

 

输入样例1

　　grid = [
　　　　　　["1","1","1","1","0"],
　　　　　　["1","1","0","1","0"],
　　　　　　["1","1","0","0","0"],
　　　　　　["0","0","0","0","0"]
　　　　  ]

输出样例1

　　1

 

输入样例2

　　grid = [
　　　　　　["1","1","0","0","0"],
　　　　　　["1","1","0","0","0"],
　　　　　　["0","0","1","0","0"],
　　　　　　["0","0","0","1","1"]
　　　　  ]

输出样例2

　　3

 

代码:

#include<iostream>
#include<vector>
#include<deque>
using namespace std;


class Solution {
public:

    void func(vector<vector<char>>& grid, int i, int j)
    {
        int m = grid.size();
        int n = grid[0].size();
        flag[i][j] = 1;
        
        if (i - 1 >= 0 && grid[i - 1][j] == '1' && !flag[i-1][j])
        {
            func(grid, i - 1, j);
        }
        if (j - 1 >= 0 && grid[i][j-1] == '1' && !flag[i][j - 1])
        {
            func(grid, i, j - 1);
        }
        if (i + 1 < m && grid[i + 1][j] == '1' && !flag[i + 1][j])
        {
            func(grid, i + 1, j);
        }
        if (j + 1 < n && grid[i][j + 1] == '1' && !flag[i][j + 1])
        {
            func(grid, i, j + 1);
        }
    }

    int numIslands(vector<vector<char>>& grid) 
    {
        int m = grid.size();
        int n = grid[0].size();
        flag = vector<vector<int>>(m, vector<int>(n, 0));
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (!flag[i][j] && grid[i][j] == '1')
                {
                    func(grid,i,j);
                    count++;
                }
            }
        }
        return count;
    }

    vector<vector<int>> flag;
    int count = 0;

};