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package class_2022_11_3_week;
// n对情侣坐在连续排列的 2n 个座位上,想要牵到对方的手
// 人和座位由一个整数数组 row 表示,其中 row[i] 是坐在第 i 个座位上的人的ID
// 情侣们按顺序编号,第一对是 (0, 1),第二对是 (2, 3),以此类推,最后一对是 (2n-2, 2n-1)
// 返回 最少交换座位的次数,以便每对情侣可以并肩坐在一起
// 每次交换可选择任意两人,让他们站起来交换座位
// 测试链接 : https://leetcode.cn/problems/couples-holding-hands/
public class Code02_CouplesHoldingHands {
public int minSwapsCouples(int[] row) {
// n人数偶数
int n = row.length;
// n/2
// 0 1 -> 0 0
// 4 5 -> 2 2
UnionFind uf = new UnionFind(n / 2);
for (int i = 0; i < n; i += 2) {
uf.union(row[i] / 2, row[i + 1] / 2);
}
return n / 2 - uf.sets();
}
public static class UnionFind {
public int[] father;
public int[] size;
public int[] help;
public int sets;
public UnionFind(int n) {
father = new int[n];
size = new int[n];
help = new int[n];
for (int i = 0; i < n; i++) {
father[i] = i;
size[i] = 1;
}
sets = n;
}
private int find(int i) {
int hi = 0;
while (i != father[i]) {
help[hi++] = i;
i = father[i];
}
while (hi != 0) {
father[help[--hi]] = i;
}
return i;
}
public void union(int i, int j) {
int fi = find(i);
int fj = find(j);
if (fi != fj) {
if (size[fi] >= size[fj]) {
father[fj] = fi;
size[fi] += size[fj];
} else {
father[fi] = fj;
size[fj] += size[fi];
}
sets--;
}
}
public int sets() {
return sets;
}
}
}