查并集

Posted 2021-03-06 weiweng

独立岛屿数

题目来源

LeetCode 200. Number of Islands

解题思路

• 暴力:遍历+dfs/bsf
• 查并集

精简解题

//解法一 dfs
func numIslands(grid [][]byte) int {
    ret := 0
    if len(grid) == 0 {
        return ret
    }
    m, n := len(grid), len(grid[0])
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if grid[i][j] == '1' {
                dfs(grid, i, j)
                ret++
            }
        }
    }
    return ret
}

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

//解法二 查并集
var match []int

func find(x int) int {
    if x != match[x] {
        match[x] = find(match[x])
    }
    return match[x]
}
func numIslands(grid [][]byte) int {
    m := len(grid)
    if m == 0 {
        return 0
    }
    n := len(grid[0])
    match = make([]int, m*n)
    for i := 0; i < m*n; i++ {
        match[i] = i
    }
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            x := i*n + j // 当前坐标
            if grid[i][j] == '0' {
                match[x] = -1
                continue
            }
            // 数字x和改行向上的数字比较
            if i != 0 && grid[i-1][j] == '1' {
                y := (i-1)*n + j
                match[find(x)] = find(y)
            }
            // 数字x和该列向左的数字比较
            if j != 0 && grid[i][j-1] == '1' {
                y := i*n + j - 1
                match[find(x)] = find(y)
            }
        }
    }
    res := 0
    mp := make(map[int]int)
    for i := 0; i < m*n; i++ {
        if match[i] == -1 {
            continue
        }
        y := find(i)
        if mp[y] != 1 {
            res++
            mp[y] = 1
        }
    }
    return res
}