[游戏开发]#952 Largest Component Size by Common Factor

Description

You are given an integer array of unique positive integers nums. Consider the following graph:

There are nums.length nodes, labeled nums[0] to nums[nums.length - 1],
There is an undirected edge between nums[i] and nums[j] if nums[i] and nums[j] share a common factor greater than 1.
Return the size of the largest connected component in the graph.

Examples

Example 1:

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

Example 2:

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

Example 3:

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

Constraints:

1 <= nums.length <= $2 * $10^4$
1 <= nums[i] <= $10^5$
All the values of nums are unique.

思路

其实我看到的时候觉得要用并查集，但我按照 $O(n^2)$的时间复杂度去计算两两数值之间是否互素，虽然通过先执行findParent来提前剪枝，但还是会TTL

但时间复杂度还是太高了，去看了下discussion，应该的方法是找到nums中最大的值，然后建立 [0, 这个值] 的图，再通过遍历找到符合nums的数字，计算最终结果。

但我觉得两种方式还是各有优劣，就都放上来吧

代码

class Solution {
    // 辗转相除法判断互素 (最后return的a就是最大公因数)
    public boolean gcd(int a, int b) {
        if (b == 0)
            return a == 1;
        return gcd(b, a % b);
    }
    
    public int findParent(int[] parents, int i) {
        if (parents[i] != i)
            parents[i] = findParent(parents, parents[i]);
        return parents[i];
    }
    
    public void union(int[] parents, int[] rank, int i, int j) {
        int parentI = findParent(parents, i);
        int parentJ = findParent(parents, j);
        int rankI = rank[parentI];
        int rankJ = rank[parentJ];
        
        if (parentI == parentJ)
            return;
        
        if (rankI > rankJ)
            parents[parentJ] = parentI;
        if (rankI < rankJ)
            parents[parentI] = parentJ;
        if (rankI == rankJ) {
            parents[parentJ] = parentI;
            rank[parentI] ++;
        }
    }
    
    public int largestComponentSize(int[] nums) {
        int[] parents = new int[nums.length];
        int[] rank = new int[nums.length];
        
        for (int i = 0; i < nums.length; i++)
            parents[i] = i;
        
        for (int i = 0; i < nums.length - 1; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (findParent(parents, i) != findParent(parents, j) && !gcd(nums[i], nums[j])) {
                    union(parents, rank, i, j);
                }
            }
        }
        
        int[] counts = new int[nums.length];
        int max = 0;
        
        for (int i = 0; i < nums.length; i++) {
            counts[findParent(parents, i)] = counts[parents[i]] + 1;
            max = Math.max(counts[parents[i]], max);
        }
        
        return max;
    }
}

class Solution {
    
    int[] p;
    
    int get(int x){
        return p[x]==x ? x : (p[x]=get(p[x])) ;
    }
    
    void add(int x, int y){
        p[get(x)]=p[get(y)];
    }
    
    public int largestComponentSize(int[] A) {
        int mx=0;
        for(int i:A) mx=Math.max(i, mx);
        p=new int[mx+1];
        for(int i=1;i<=mx;++i) p[i]=i;
        for(int i:A)
            for(int j=(int)Math.sqrt(i);j>=2;--j)
                if(i%j==0){
                    add(i, j);
                    add(i, i/j);
                }
        Map<Integer, Integer> map=new HashMap<Integer, Integer>();
        int res=0;
        for(int i:A){
            int j=get(i);
            if(!map.containsKey(j)) map.put(j, 1);
            else map.put(j, map.get(j)+1);
            res=Math.max(res, map.get(j));
        }
        return res;
    }
}

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