648. Replace Words

Medium

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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 the ​​root​​​ 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"
typedef struct Trie_node{
    bool exist;
    struct Trie_node* next[26];
    Trie_node():exist(false){
        memset(next,0,sizeof(Trie_node*) * 26);
    }
}TrieNode,*Trie;
class Solution {
public:
    string replaceWords(vector<string>& dict, string sentence) {
        string Result;
        if(sentence.length() == 0){
            return Result;
        }
        Trie root = new TrieNode();
        BuildTrieTree(root,dict);

        stringstream ss(sentence);
        string temp;
        ss >> temp;
        Result += FindStr(root,temp);
        while(ss >> temp){
            Result = Result + " " + FindStr(root,temp);
        }
        return Result;
    }
    void BuildTrieTree(Trie root,vector<string>& dicts){
        for(auto dict : dicts){
            Trie p = root;
            for(int i = 0;i < dict.length();i ++){
                int index = dict[i] - 'a';
                if(p->next[index] == NULL){
                    p->next[index] = new Trie_node();
                }
                p = p->next[index];
                if(i == dict.length() - 1){
                    p->exist = true;
                }
            }
        }
    }
    string FindStr(Trie root,const string& str){
        Trie p = root;
        for(int i = 0;i < str.length();i ++){
            int index = str[i] - 'a';
            if(p->next[index]){
                if(p->next[index]->exist){
                    return str.substr(0,i + 1);
                }
                p = p->next[index];
            }
            else{
                return str;
            }
        }
        return str;
    }
};