题目链接：https://leetcode-cn.com/problems/baby-names-lcci/
分析题意可以知道这是个并查集，新增一个size数目来维护集合数量大小就可以了(size数组只有当是父节点时才有意义)
思路：将synonyms中的相同的名字合并即可，但是names的处理就比较麻烦，对于John(15)这样一个字符串，要分离出John和15这两个字符串。同时还有一个要注意点，并查集merge的时候要注意如果find的值相同(是同一个集合的)，是不需要重复加的。

class Solution:def trulyMostPopular(self, names: List[str], synonyms: List[str]) -> List[str]:# initself.p = list(range(0, len(names)+100, 1))self.size = [0] * (len(names)+100);self.name_index_dic = {}for i in range(len(names)):name = names[i]index= name.find('(')name_size = name[index+1:-1]name = name[:index]self.name_index_dic[name] = iself.size[self.name_index_dic[name]] = int(name_size)names[i] = nameidx = len(names)# 合并for synonym in synonyms:index = synonym.find(',')name1 = synonym[1:index]name2 = synonym[index+1:-1]name1_id = self.name_index_dic.get(synonym[1:index])name2_id = self.name_index_dic.get(synonym[index+1:-1])if name1_id == None:self.name_index_dic[name1] = idxnames.append(name1)name1_id = idxidx += 1if name2_id == None:self.name_index_dic[name2] = idxnames.append(name2)name2_id = idxidx += 1# print(str(name1_id)+" "+str(name2_id))self.__merge(name1_id, name2_id)# 路径压缩for i in range(len(names)):self.__find(i)# 得到字典序最小的名字index_name_dict = {}for i in range(len(names)):if index_name_dict.__contains__(self.p[i]):now = index_name_dict.get(self.p[i])if names[i] < now:index_name_dict[self.p[i]] = names[i]else:index_name_dict[self.p[i]] = names[i];# 组装答案ans = []for key in index_name_dict:s = index_name_dict[key] + "(" + str(self.size[key]) + ")"ans.append(s)# print(self.p)return ansp = []size = []name_index_dic = {}def __find(self, x : int):if x != self.p[x]:self.p[x] = self.__find(self.p[x])return self.p[x]def __merge(self, a : int, b : int):a = self.__find(a)b = self.__find(b)if a != b:self.p[a] = b;self.size[b] += self.size[a]