LeetCode 648. Replace Words 字典树练习

In English, we have a concept called root, which can be followed by some other words to form another longer word - let's call this word successor. For example, the root an, followed by other, which can form another word another.

Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor in the sentence with theroot forming it. If a successor has many roots can form it, replace it with the root with the shortest length.

You need to output the sentence after the replacement.

Example 1:

Input: dict = ["cat", "bat", "rat"]sentence = "the cattle was rattled by the battery"Output: "the cat was rat by the bat"

题意：用字典中存在的前缀代替句子中的单词，若有多个前缀可以表示单词，则选择最短的一个

用基础字典树即可解决，将字典中的前缀字符串存入字典树中，遍历每个句子的单词，找到最短的前缀返回即可

字典树结构：

struct Trie{int val;//表示此位置是否有前缀结束vector<Trie*> child;Trie():child(vector<Trie*>(26,NULL)),val(-1){}};

首先是字典树的创建（插入）过程，题目只要返回最短的前缀字符串，在插入时对于每一个字符串的最后一个字符节点的val设置为0表示此位置为结束位置。

void insert(string s,Trie* root){for(int i=0;i<s.size();i++){int num = s[i]-'a';if(root->child[num] == NULL){root->child[num] = new Trie();}root = root->child[num];        if(i!=s.size()-1&&root->val!=0)                root->val = 1;        else root->val = 0;//若此位置为结束位置，将val设置为0}}   

查找时，遇到val为0的字符节点就可以返回，这样就能保证返回的前缀为最短的，若字典树中不存在此单词的前缀，那么查找停止位置的val不等于0；

    string search(string s,Trie* root){string res = "";for(int i=0;i<s.size();i++){int num = s[i] -'a';if(root->child[num] == NULL)break;        root = root->child[num];    res += s[i];        if(root->val == 0)            return res;}        if(root->val == 0)        return res;        else return s;}

在分割句子时可以使用强大的stringstream，将sentence先读入字符串流中，以空格为标示分割句子，超级方便，字符串分割利器，还能类型转化等等：

string replaceWords(vector<string>& dict, string sentence) {        Trie *root = new Trie();        string res = "";        for(string s:dict)            insert(s,root);        stringstream ss(sentence);        string s;        int sign = 1;        while(ss>>s){            string temp = search(s,root);            if(sign!=1)                res += " ";            if(temp.empty())                res +=s;            else res +=temp;            sign++;        }        return res;    }