LeetCode-200.岛屿数量

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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

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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
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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'

题解

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func numIslands(_ grid: [[Character]]) -> Int {
  guard !grid.isEmpty else {
    return 0
  }
  var numbers = grid
  var count = 0
  
  let row = numbers.count
  let col = numbers[0].count
  
  func dfs( _ i: Int, _ j: Int) {
    // i/j到达边界或者(i,j)位置不为“1”，停止搜索
    if i < 0 || i > row - 1 || j < 0 || j > col - 1 || numbers[i][j] != "1" {
      return
    }
    numbers[i][j] = "0" // 将(i, j)位置重置为“0”
    dfs(i, j - 1) // 左
    dfs(i, j + 1) // 右
    dfs(i - 1, j) // 上
    dfs(i + 1, j) // 下
  }
  
  for i in 0 ..< row {
    for j in 0 ..< col {
      if numbers[i][j] == "1" {
        count += 1
        dfs(i, j)
      }
    }
  }
  
  return count
}