leecode 面试题 17.07 婴儿名字

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]:
        # init
        self.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] = i
            self.size[self.name_index_dic[name]] = int(name_size)
            names[i] = name

        idx = 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] = idx
                names.append(name1)
                name1_id = idx
                idx += 1
            if name2_id == None:
                self.name_index_dic[name2] = idx
                names.append(name2)
                name2_id = idx
                idx += 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 ans

    p = []
    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]