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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
}

LC-岛屿数量

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原文地址：https://www.cnblogs.com/wdy1184/p/11599461.html