力扣（LeetCode）200. 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。

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 {
private:
    int dir[4][2] = {{0,1}, {1,0}, {-1,0}, {0,-1}};//右上左下，四个方向
    void dfs(vector<vector<char>>& grid, vector<vector<bool>>& vis, int x, int y)
    {
        for(int i = 0; i < 4; i++)//遍历四个方向
        {
            int nextx = x + dir[i][0];
            int nexty = y + dir[i][1];
            //判断边界条件
            if(nextx < 0 || nexty < 0 || nextx >= grid.size() || nexty >= grid[0].size())
                continue;
            if(!vis[nextx][nexty] && grid[nextx][nexty] == '1')
            {
                vis[nextx][nexty] = true;
                dfs(grid, vis, nextx, nexty);
            }
        }
    }
public:
    int numIslands(vector<vector<char>>& grid) {
        int n = grid.size(), m = grid[0].size();

        //初始化vis为一个n行m列，且元素均为false的二维数组
        vector<vector<bool>> vis = vector<vector<bool>>(n, vector<bool>(m, false));

        int count = 0;//岛屿数量
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                //两重循环遍历二维网格grid
                if(!vis[i][j] && grid[i][j] == '1')
                //如果未被访问且为陆地，则以该点为起点，将周围陆地均改为true
                {
                    count++;

                    //遍历周围陆地
                    dfs(grid, vis, i, j); 
                }
            }
        }
        return count;
    }
};

题目链接：https://leetcode.cn/problems/number-of-islands/