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'
func numIslands(grid [][]byte) int {
    
}

func numIslands(grid [][]byte) int {
	ans := 0
	var dfs func(row, col int)
	dfs = func(i, j int) { // 深度优先 尽可能往里面走
		grid[i][j] = '0'
		// 判断四周的
		// 上
		if i > 0 && grid[i-1][j] == '1' {
			dfs(i-1, j)
		}
		// 左
		if j > 0 && grid[i][j-1] == '1' {
			dfs(i, j-1)
		}
		// 右
		if i < len(grid)-1 && grid[i+1][j] == '1' {
			dfs(i+1, j)
		}
		// 下
		if j < len(grid[i])-1 && grid[i][j+1] == '1' {
			dfs(i, j+1)
		}
	}
	for i := 0; i < len(grid); i++ {
		for j := 0; j < len(grid[i]); j++ {
			if grid[i][j] == '1' {
				ans++
				dfs(i, j)
			}
		}
	}
	return ans
}