題目

https://leetcode.com/problems/accounts-merge/

題解

HashMap 版的并查集。參考了：leetcode 684. Redundant Connection | 684. 冗余連接（并查集）

class Solution {public List<List<String>> accountsMerge(List<List<String>> accounts) {Map<String, String> parent = new HashMap<>();Map<String, String> name = new HashMap<>();Map<String, Integer> size = new HashMap<>();// initfor (List<String> list : accounts) {name.put(list.get(1), list.get(0));parent.put(list.get(1), list.get(1));size.put(list.get(1), 1);for (int i = 2; i < list.size(); i++) {name.put(list.get(i), list.get(0));parent.put(list.get(i), list.get(i));size.put(list.get(i), 1);}}// union findfor (List<String> list : accounts) {for (int i = 2; i < list.size(); i++) {String r0 = findRoot(parent, list.get(1)); // r0會被修改，不能放在外層不變！！String r1 = findRoot(parent, list.get(i));if (r0.equals(r1)) continue;if (size.get(r0) > size.get(r1)) {parent.put(r1, r0);} else {parent.put(r0, r1);}size.put(r0, size.get(r0) + size.get(r1));size.put(r1, size.get(r0) + size.get(r1));}}HashMap<String, Set<String>> resMap = new HashMap<>();for (List<String> list : accounts) {for (int i = 1; i < list.size(); i++) {String root = findRoot(parent, list.get(i));resMap.computeIfAbsent(root, k -> new TreeSet<>());resMap.get(root).add(list.get(i));}}List<List<String>> result = new ArrayList<>();for (String root : resMap.keySet()) {List<String> list = new ArrayList<>();list.add(name.get(root));list.addAll(resMap.get(root));result.add(list);}return result;}// 從s開始一直往上找，直到不能再往上，找到代表節點，返回public String findRoot(Map<String, String> parent, String s) {String cur = s;while (!cur.equals(parent.get(cur))) {cur = parent.get(cur);}while (!s.equals(parent.get(s))) { // 路徑壓縮String next = parent.get(s);parent.put(s, cur);s = next;}return cur;} }

總結

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