Hard Level

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：

$m == grid.length$
$n == grid[i].length$
$1 <= m, n <= 300$
$grid[i][j] 的值为 '0' 或 '1'$

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Flood Filled

class Solution {
public:
    vector<vector<char>> g;

    int numIslands(vector<vector<char>>& grid) {
        g = grid;
        int cnt = 0;
        for(int i = 0; i < g.size(); i++) {
            for(int j = 0; j < g[0].size(); j++) {
                if(g[i][j] == '1') {
                    dfs(i, j);
                    cnt++;
                }
            }
        }
        return cnt;
    }

    int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

    void dfs(int x, int y) {
        g[x][y] = 0;
        for(int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if(a >= 0 && a < g.size() && b >= 0 && b < g[0].size() && g[a][b] == '1') {
                dfs(a, b);
            }
        }
    }
};