200. 岛屿数量

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

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’

代码 

const (
	valid   byte = '1'
	inValid byte = '0'
)

func numIslands(grid [][]byte) int {
	copyGrid := make([][]byte, 0)
	copyGrid = append(copyGrid, grid...)
	var count int
	for i := 0; i < len(copyGrid); i++ {
		for j := 0; j < len(copyGrid[i]); j++ {
			if copyGrid[i][j] == valid {
				mark(copyGrid, i, j)
				count++
			}
		}
	}
	return count
}

func mark(copyGrid [][]byte, i, j int) {
	if i < 0 || i >= len(copyGrid) || j < 0 || j >= len(copyGrid[i]) {
		return
	}
	if copyGrid[i][j] == inValid {
		return
	}
	copyGrid[i][j] = inValid
	mark(copyGrid, i, j+1)
	mark(copyGrid, i, j-1)
	mark(copyGrid, i+1, j)
	mark(copyGrid, i-1, j)
}