LeetCode-200. Number of Islands

发表于 分类于

问题描述

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

1
2
3
4
5
6
7
Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

1
2
3
4
5
6
7
Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

Constraints:

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

解答

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
/*
 * @Description: 200. Number of Islands
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * @param {character[][]} grid
 * @return {number}
 */
const numIslands = function (grid) {
  if (grid === null || grid.length === 0) {
    return 0
  }
  let count = 0
  const row = grid.length
  const col = grid[0].length

  const dfs = (grid, i, j) => {
    if (i < 0 || i > row - 1 || j < 0 || j > col - 1 || grid[i][j] !== '1') {
      return
    }
    grid[i][j] = '2'
    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 < row; i++) {
    for (let j = 0; j < col; j++) {
      if (grid[i][j] === '1') {
        count++
        dfs(grid, i, j)
      }
    }
  }
  return count
}