[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' 
// 
// Related Topics 深度优先搜索 广度优先搜索 并查集 
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/*
* 200 岛屿数量
* 2021-06-14 13:07:54
* @author oxygenbytes
*/ 
#include "leetcode.h" 
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int 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;
    }
    int dx[4] = {-1, 0, 1, 0};
    int dy[4] = {0, 1, 0, -1};
    void dfs(vector<vector<char>>& grid, int x,int y){
        grid[x][y] = '0';
        for(int i = 0;i < 4;i++){
            int a = dx[i] + x;
            int b = dy[i] + y;
            if(a < 0 || a >= grid.size() || b < 0 || b >= grid[0].size() || grid[a][b] == '0')
                continue;
            dfs(grid, a, b);
        }
    }
};
//leetcode submit region end(Prohibit modification and deletion)