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：

  1. 输入：grid = [
  2.   ["1","1","1","1","0"],
  3.   ["1","1","0","1","0"],
  4.   ["1","1","0","0","0"],
  5.   ["0","0","0","0","0"]
  6. ]
  7. 输出：1

示例 2：

  1. 输入：grid = [
  2.   ["1","1","0","0","0"],
  3.   ["1","1","0","0","0"],
  4.   ["0","0","1","0","0"],
  5.   ["0","0","0","1","1"]
  6. ]
  7. 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'

    解法一：DFS

    岛屿问题标准模版 ```go var pairs = []struct{ x, y int }{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}

func numIslands(grid [][]byte) int { m := len(grid) n := len(grid[0]) var res int for i := 0; i < m; i++ { for j := 0; j < n; j++ { if grid[i][j] == ‘1’ { dfs(i, j, grid) res++ } } } return res }

func dfs(i, j int, grid [][]byte) { if i < 0 || j < 0 || i >= len(grid) || j >= len(grid[0]) || grid[i][j] != ‘1’ { return } grid[i][j] = ‘2’ for _, pair := range pairs { dfs(i+pair.x, j+pair.y, grid) } return } ```