岛屿数量

岛屿数量

​ 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'

题解：

​ dfs 遍历即可

func numIslands(grid [][]byte) int {
	m, n := len(grid), len(grid[0])
	vis := make([][]bool, m)
	for i := range vis {
		vis[i] = make([]bool, n)
	}
	dirs := [][]int{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
	ans := 0

    var dfs func(x, y int)
	dfs = func(x, y int) {
		vis[x][y] = true
		for _, dir := range dirs {
			nextX, nextY := x+dir[0], y+dir[1]
			if nextX >= 0 && nextX < m && nextY >= 0 && nextY < n && !vis[nextX][nextY] && grid[nextX][nextY] == '1' {
				dfs(nextX, nextY)
			}
		}
	}

	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if vis[i][j] == false && grid[i][j] == '1' {
				ans++
				dfs(i, j)
			}
		}
	}
	
	return ans
}