200.岛屿数量

1、题干

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'

解题思路

• 遍历矩阵grid，以grid[i][j]为起点往上下左右4个方向进行深度遍历
• 深度遍历过程中需要注意2点
  • 遍历过的节点修改为0，即grid[i][j] = '0'，以免重复遍历
  • grid[i][j]必须在矩阵中且grid[i][j] === '1'才累加岛屿数量

代码

var numIslands = function (grid) {
    const dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]];
    function dfs(i, j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] === '0') return false;
        grid[i][j] = '0';
        for (const [di, dj] of dirs) dfs(i + di, j + dj);
        return true;
    }

    let count = 0;
    for (let i = 0; i < grid.length; i++) {
        for (let j = 0; j < grid[i].length; j++) {
            if (dfs(i, j)) count++;
        }
    }

    return count;
};