200. [✔][M]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：

岛屿问题

// @lc code=start
function numIslands(grid: string[][]): number {
    let m = grid.length;
    let n = grid[0].length;

    let res = 0;

    const dfs = (grid: string[][], i: number, j: number) => {
        if (i < 0 || i >= m || j < 0 || j >= n) {
            return;
        }
        if (grid[i][j] === "0") {
            return;
        }

        grid[i][j] = "0";
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);

    }

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (grid[i][j] === "1") {
                res++;
                dfs(grid, i, j);//找到1个”1“，就把这个1所在的岛屿变成海水
            }
        }
    }

    return res;
};
// @lc code=end