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。

示例 1：
输入：

  1. 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. ]

输出：1

示例 2：
输入：

  1. 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. ]

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’

    方案（回溯）

    ```go func numIslands(grid [][]byte) int { m := len(grid) n := len(grid[0])

    ans := 0

    for i, row := range grid {

      for j, each := range row {
      if each == "0"[0] || each == 2 {
          continue
      }

      ans += 1
      dfs(&grid, i, j, m, n)
  }

    }

    return ans }

func dfs(grid [][]byte, i, j, m, n int) { (grid)[i][j] = byte(2) for _, ij := range around(i, j, m, n) { if (grid)[ij[0]][ij[1]] == 2 || (grid)[ij[0]][ij[1]] == “0”[0] { continue }

    dfs(grid, ij[0], ij[1], m, n)
}

}

func around(i, j, m, n int) [][2]int { ret := [][2]int{}

for _, each := range [][2]int{
    {i-1, j},
    {i+1, j},
    {i, j-1},
    {i, j+1},
} {
    if 0 <= each[0] && each[0] < m && 0 <= each[1] && each[1] < n {
        ret = append(ret, each)
    }
}

return ret

} ```

原文

https://leetcode-cn.com/problems/number-of-islands/