LC-岛屿数量

岛屿数量

Date Created: Sep 22, 2019 11:28 PM
Last Edited Time: Sep 27, 2019 5:39 PM
显示完成时间: Sep 23, 2019
状态: 已完成
真实完成时间: Sep 23, 2019
类型: leetcode

解法一

深度优先遍历，并且将遍历过的数据进行标记

func numIslands(grid [][]byte) int {
	if len(grid) == 0 || len(grid[0]) == 0{
		return 0
	}

	count := 0
	for i, iNum := range grid {
		for j, _ := range iNum {
			if grid[i][j] == '1' {
				count ++
				dfs(grid, i, j)
			}
		}
	}

	return count
}

func dfs(grid [][]byte, i int, j int)  {
	grid[i][j] = '0'
	if i - 1 >= 0 && grid[i-1][j] == '1' {
		dfs(grid, i-1, j)
	}
	if j - 1 >= 0 && grid[i][j - 1] == '1' {
		dfs(grid, i, j-1)
	}
	if i + 1 < len(grid) && grid[i+1][j] == '1' {
		dfs(grid, i+1, j)
	}
	if j+1 < len(grid[0]) && grid[i][j+1] == '1' {
		dfs(grid, i, j+1)
	}
}

解法二

广度优先遍历，将遍历过的数据进行标记

func numIslands(grid [][]byte) int {
	if len(grid) == 0 || len(grid[0]) == 0 {
		return 0
	}
	isTouched := make([][]bool, len(grid))
	for i, _ := range isTouched {
		isTouched[i] = make([]bool, len(grid[0]))
	}

	count := 0
	queue := list.New()
	for i := 0; i < len(grid); i++ {
		for j := 0; j < len(grid[0]); j++ {
			if grid[i][j] == '1' {
				count++
				queue.PushBack(pos{i, j})
				for queue.Len() > 0 {
					ele := queue.Front()
					v := queue.Remove(ele).(pos)
					grid[v.i][v.j] = '0'
					if v.i-1 >= 0 && !isTouched[v.i-1][v.j] && grid[v.i-1][v.j] == '1' {
						queue.PushBack(pos{v.i - 1, v.j})
						isTouched[v.i-1][v.j] = true
					}
					if v.j+1 < len(grid[0]) && !isTouched[v.i][v.j+1] && grid[v.i][v.j+1] == '1' {
						queue.PushBack(pos{v.i, v.j + 1})
						isTouched[v.i][v.j+1] = true
					}
					if v.i+1 < len(grid) && !isTouched[v.i+1][v.j] && grid[v.i+1][v.j] == '1' {
						queue.PushBack(pos{v.i + 1, v.j})
						isTouched[v.i+1][v.j] = true
					}
					if v.j-1 >= 0 && !isTouched[v.i][v.j-1] && grid[v.i][v.j-1] == '1' {
						queue.PushBack(pos{v.i, v.j - 1})
						isTouched[v.i][v.j-1] = true
					}
				}
			}
		}
	}

	return count
}

type pos struct {
	i int
	j int
}