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岛屿数量

题目​

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.

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

题解​

核心思想是遍历网格, 找到一个陆地 '1' 后, 将其相邻的陆地都标记为已访问, 直到该岛屿的所有陆地都被标记.

/**
 * @param {character[][]} grid
 * @return {number}
 */
var numIslands = function (grid) {
  const m = grid.length
  const n = grid[0].length
  let count = 0

  const dfs = (i, j) => {
    if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] === '0') {
      return
    }

    grid[i][j] = '0' // 标记为已访问

    dfs(i + 1, j)
    dfs(i - 1, j)
    dfs(i, j + 1)
    dfs(i, j - 1)
  }

  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (grid[i][j] === '1') {
        count++
        dfs(i, j)
      }
    }
  }

  return count
}