200. 岛屿数量

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200. 岛屿数量

Description

Difficulty: 中等

Related Topics: 深度优先搜索, 广度优先搜索, 并查集, 数组, 矩阵

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'

Solution

Language: JavaScript

/**
 * @param {character[][]} grid
 * @return {number}
 */
var numIslands = function(grid) {
    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++
            }
        }
    }

    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)
    }
    return count
}