200\. 岛屿数量

# 200. 岛屿数量 (opens new window)

# Description

Difficulty: 中等

Related Topics: 深度优先搜索 (opens new window), 广度优先搜索 (opens new window), 并查集 (opens new window), 数组 (opens new window), 矩阵 (opens new window)

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
1
2
3
4
5
6
7

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
1
2
3
4
5
6
7

提示:

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

# Solution

Language: JavaScript

var numIslands = function(grid) {
    function dfs(x, y) {
        grid[x][y] = 0
        if(grid[x - 1] && grid[x - 1][y] == 1) dfs(x-1, y)
        if(grid[x + 1] && grid[x + 1][y] == 1) dfs(x+1, y)
        if(grid[x][y - 1] == 1) dfs(x, y - 1)
        if(grid[x][y + 1] == 1) dfs(x, y + 1)
    }
    let count = 0
    for(let i = 0; i < grid.length; i++) {
        for(let j = 0; j< grid[0].length; j++) {
            if(grid[i][j] == 1) {
                dfs(i, j)
                count++
            }
        }
    }
    return count
};

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
上次更新: 2022/8/23 下午4:17:54