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package class20;
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import java.util.HashMap;
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// 本题为leetcode原题
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// 测试链接：https://leetcode.com/problems/largest-component-size-by-common-factor/
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// 方法1会超时，但是方法2直接通过
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public class Code02_LargestComponentSizebyCommonFactor {
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public static int largestComponentSize1(int[] arr) {
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int N = arr.length;
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UnionFind set = new UnionFind(N);
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for (int i = 0; i < N; i++) {
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for (int j = i + 1; j < N; j++) {
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if (gcd(arr[i], arr[j]) != 1) {
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set.union(i, j);
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}
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}
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}
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return set.maxSize();
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}
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public static int largestComponentSize2(int[] arr) {
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int N = arr.length;
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// arr中，N个位置，在并查集初始时，每个位置自己是一个集合
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UnionFind unionFind = new UnionFind(N);
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// key 某个因子 value 哪个位置拥有这个因子
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HashMap<Integer, Integer> fatorsMap = new HashMap<>();
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for (int i = 0; i < N; i++) {
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int num = arr[i];
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// 求出根号N， -> limit
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int limit = (int) Math.sqrt(num);
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for (int j = 1; j <= limit; j++) {
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if (num % j == 0) {
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if (j != 1) {
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if (!fatorsMap.containsKey(j)) {
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fatorsMap.put(j, i);
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} else {
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unionFind.union(fatorsMap.get(j), i);
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}
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}
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int other = num / j;
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if (other != 1) {
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if (!fatorsMap.containsKey(other)) {
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fatorsMap.put(other, i);
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} else {
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unionFind.union(fatorsMap.get(other), i);
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}
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}
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}
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}
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}
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return unionFind.maxSize();
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}
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// O(1)
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// m,n 要是正数，不能有任何一个等于0
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public static int gcd(int a, int b) {
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return b == 0 ? a : gcd(b, a % b);
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}
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public static class UnionFind {
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private int[] parents;
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private int[] sizes;
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private int[] help;
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public UnionFind(int N) {
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parents = new int[N];
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sizes = new int[N];
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help = new int[N];
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for (int i = 0; i < N; i++) {
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parents[i] = i;
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sizes[i] = 1;
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}
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}
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public int maxSize() {
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int ans = 0;
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for (int size : sizes) {
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ans = Math.max(ans, size);
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}
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return ans;
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}
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private int find(int i) {
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int hi = 0;
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while (i != parents[i]) {
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help[hi++] = i;
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i = parents[i];
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}
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for (hi--; hi >= 0; hi--) {
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parents[help[hi]] = i;
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}
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return i;
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}
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public void union(int i, int j) {
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int f1 = find(i);
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int f2 = find(j);
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if (f1 != f2) {
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int big = sizes[f1] >= sizes[f2] ? f1 : f2;
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int small = big == f1 ? f2 : f1;
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parents[small] = big;
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sizes[big] = sizes[f1] + sizes[f2];
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}
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}
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}
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}
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