200. 岛屿数量

mid

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

DFS

我们可以将二维网格看成一个无向图，竖直或水平相邻的 1 之间有边相连。

为了求出岛屿的数量，我们可以扫描整个二维网格。如果一个位置为 1，则以其为起始节点开始进行深度优先搜索。在深度优先搜索的过程中，每个搜索到的 1 都会被重新标记为 0。（或者用一个 visited 数组标记到达过的陆地）

最终岛屿的数量就是我们进行深度优先搜索的次数。

// with visited
func numIslands(grid [][]byte) int {
	rows, cols := len(grid), len(grid[0])
	visited := make([][]bool, rows)
	for i, _ := range visited {
		visited[i] = make([]bool, cols)
	}
	var explore func(x, y int)
	explore = func(x, y int) {
		if x < 0 || x > rows-1 || y < 0 || y > cols-1 || grid[x][y] == '0' || visited[x][y] {
			return
		}
		visited[x][y] = true
		explore(x+1, y)
		explore(x, y+1)
		explore(x-1, y)
		explore(x, y-1)
	}
	count := 0
	for x := 0; x < rows; x++ {
		for y := 0; y < cols; y++ {
			if grid[x][y] == '0' || visited[x][y] == true {
				continue
			}
			count++
			explore(x, y)

		}
	}
	return count
}

// without visited
func numIslands(grid [][]byte) int {
	rows, cols := len(grid), len(grid[0])
	var explore func(x, y int)
	explore = func(x, y int) {
		if x < 0 || x > rows-1 || y < 0 || y > cols-1 || grid[x][y] == '0' {
			return
		}
		grid[x][y] = '0'
		explore(x+1, y)
		explore(x, y+1)
		explore(x-1, y)
		explore(x, y-1)
	}
	count := 0
	for x := 0; x < rows; x++ {
		for y := 0; y < cols; y++ {
			if grid[x][y] == '0' {
				continue
			}
			count++
			explore(x, y)

		}
	}
	return count
}