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80 lines
1.8 KiB
80 lines
1.8 KiB
package 第03期.mca_06;
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// n对情侣坐在连续排列的 2n 个座位上，想要牵到对方的手
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// 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
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// 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)
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// 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
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// 每次交换可选择任意两人，让他们站起来交换座位
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// 测试链接 : https://leetcode.cn/problems/couples-holding-hands/
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public class Code02_CouplesHoldingHands {
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public int minSwapsCouples(int[] row) {
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// 数字几个 2 * n 8 4 16 8
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// 0 1 2 3 4 5 v
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// 0 1 2 v/2
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// n : 人数
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// 情侣组 : n / 2
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int n = row.length;
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UnionFind uf = new UnionFind(n / 2);
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for (int i = 0; i < n; i += 2) {
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// 左 右 左 右
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// 0 1 2 3 4
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uf.union(row[i] / 2, row[i + 1] / 2);
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}
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// n / 2 -
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return n / 2 - uf.sets();
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}
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public static class UnionFind {
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public int[] father;
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public int[] size;
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public int[] help;
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// 当前有多少个集合？
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public int sets;
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public UnionFind(int n) {
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father = new int[n];
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size = 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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father[i] = i;
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size[i] = 1;
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}
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sets = n;
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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 != father[i]) {
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help[hi++] = i;
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i = father[i];
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}
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while (hi != 0) {
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father[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 fi = find(i);
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int fj = find(j);
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if (fi != fj) {
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if (size[fi] >= size[fj]) {
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father[fj] = fi;
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size[fi] += size[fj];
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} else {
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father[fi] = fj;
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size[fj] += size[fi];
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}
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sets--;
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}
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}
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public int sets() {
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return sets;
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}
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}
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}
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