[Swift]LeetCode952. 按公因数计算最大组件大小 | Largest Component Size by Common Factor

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Given a non-empty array of unique positive integers A, consider the following graph:

• There are A.length nodes, labelled A[0] to A[A.length - 1];
• There is an edge between A[i] and A[j] if and only if A[i] and A[j] share a common factor greater than 1.

Return the size of the largest connected component in the graph.

Example 1:

Input: [4,6,15,35]
Output: 4

Example 2:

Input: [20,50,9,63]
Output: 2

Example 3:

Input: [2,3,6,7,4,12,21,39]
Output: 8

Note:

1. 1 <= A.length <= 20000
2. 1 <= A[i] <= 100000

---

给定一个由不同正整数的组成的非空数组 A，考虑下面的图：

• 有 A.length 个节点，按从 A[0] 到 A[A.length - 1] 标记；
• 只有当 A[i] 和 A[j] 共用一个大于 1 的公因数时，A[i] 和 A[j] 之间才有一条边。

返回图中最大连通组件的大小。

示例 1：

输入：[4,6,15,35]
输出：4

示例 2：

输入：[20,50,9,63]
输出：2

示例 3：

输入：[2,3,6,7,4,12,21,39]
输出：8

提示：

1. 1 <= A.length <= 20000
2. 1 <= A[i] <= 100000

---

2676 ms 

class Solution {
    func largestComponentSize(_ A: [Int]) -> Int {
        var lpf:[Int] = enumLowestPrimeFactors(100001)
        var ds:DJSet = DJSet(100001)
        for v in A
        {
            ds.w[v] = 1
        }
        for v in A
        {
            var vv:Int = v
            while(vv > 1)
            {
                ds.union(v, lpf[vv])
                vv /= lpf[vv]
            }
        }
        var ret:Int = 0
        for i in 1...100000
        {
            if ds.upper[i] < 0
            {
                ret = max(ret, ds.w[i])
            }
        }
        return ret
    }
    
    func enumLowestPrimeFactors(_ n:Int) -> [Int]
    {
        var tot:Int = 0
        var lpf:[Int] = [Int](repeating:0,count:n + 1)
        var u:Double = Double(n + 32)
        var lu:Double = log(u)
        let num:Int = Int(u / lu + u / lu / lu * 1.5)
        var primes:[Int] = [Int](repeating:0,count:num)
        for i in 2...n
        {
            lpf[i] = i
        }
        for p in 2...n
        {
            if lpf[p] == p
            {
                primes[tot] = p
                tot += 1
            }
            var tmp:Int = 0
            var i:Int = 0
            while(i < tot && primes[i] <= lpf[p] && primes[i] * p <= n)
            {
                tmp = primes[i] * p
                lpf[tmp] = primes[i]
                i += 1
            }
        }
        return lpf
    }
}
public class DJSet
{
    var upper:[Int]
    var w:[Int]
    
    init(_ n:Int)
    {
        upper = [Int](repeating:-1,count:n)
        w = [Int](repeating:0,count:n)
    }
    
    func root(_ x:Int) -> Int
    {
        if(upper[x] < 0)
        {
            return x
        }
        else
        {
            upper[x] = root(upper[x])
            return upper[x]
        }
    }
    
    func equiv(_ x:Int,_ y:Int) -> Bool
    {
        return root(x) == root(y)
    }
    
    func union(_ x:Int,_ y:Int) -> Bool
    {
        var x:Int = root(x)
        var y:Int = root(y)
        if x != y
        {
            if upper[y] < upper[x]
            {
                var d:Int = x
                x = y
                y = d
            }
            upper[x] += upper[y]
            upper[y] = x
            w[x] += w[y]
        }
        return x == y
    }
    
    func count() -> Int
    {
        var ct:Int = 0
        for u in upper
        {
            if u < 0
            {
                ct += 1
            }
        }
        return ct
    }
}