648. Replace Words（LeetCode）

Posted 2020-10-03 无惧风云

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"

 

Note:

1. The input will only have lower-case letters.
2. 1 <= dict words number <= 1000
3. 1 <= sentence words number <= 1000
4. 1 <= root length <= 100
5. 1 <= sentence words length <= 1000

    1 class Solution {
    2 public:
    3     string replaceWords(vector<string>& dict, string sentence) {
    4         if (sentence.size() == 0 || dict.size() == 0)
    5             return sentence;
    6         stringstream ss(sentence);
    7         vector<string> vet;
    8         string str;
    9         while (ss >> str)
   10         {
   11             vet.push_back(str);
   12         }
   13         for (int i = 0; i < vet.size(); i++)
   14         {
   15             for (int j = 0; j < dict.size(); j++)
   16             {
   17                 /*cout << vet[i] << vet[i].size() << " " << dict[j] << dict[j].size() << "  22" << vet[i].substr(0, dict[j].size()) << endl;*/
   18 
   19                 if (vet[i].size() >= dict[j].size() && vet[i].substr(0, dict[j].size()) == dict[j])
   20                 {
   21                             vet[i] = dict[j];    
   22                             /*cout <<"11111111"<< vet[i] << endl;*/
   23                 }
   24             }
   25         }
   26         string str1 = "";
   27         for (int i = 0; i < vet.size(); i++)
   28         {
   29             str1 += vet[i];
   30             if (i != vet.size()-1)
   31             str1 += " ";
   32 
   33         }
   34         return str1;
   35     }
   36 };

   别人用字典树做的如下：

    1 class Solution {
    2 private:
    3     class TrieNode{
    4         public:
    5         bool eow;
    6         vector<TrieNode*> chars;
    7         TrieNode():eow(false), chars(vector<TrieNode*>(26,NULL)){
    8             
    9         }
   10     };
   11     
   12     class Trie{
   13         public:
   14         TrieNode* head;
   15         
   16         Trie():head(new TrieNode()) {}
   17         
   18         bool isWord(string str){
   19            TrieNode* cur = head;
   20             for (int i = 0; i < str.size(); i++){
   21                 char c = str[i];
   22                 int index = (c-‘a‘) %26;
   23                 if (cur->chars[index] == NULL){
   24                     return false;
   25                 }
   26                 cur = cur->chars[index];
   27                 if (index == str.size() - 1){
   28                     return cur->eow;
   29                 }
   30             }
   31         }
   32         
   33         void insertWord(string str){
   34             TrieNode* cur = head;
   35             for (int i = 0; i < str.size(); i++){
   36                 char c = str[i];
   37                 int index = (c-‘a‘) %26;
   38                 if (cur->chars[index] == NULL){
   39                     cur->chars[index] = new TrieNode();
   40                 }
   41                 cur = cur->chars[index];
   42                 if (i == str.size() - 1){
   43                     cur->eow = true;
   44                 }
   45             }
   46         }
   47         
   48         string getShortestPrefix(string str){
   49             TrieNode* cur = head;
   50             for (int i = 0; i < str.size(); i++){
   51                 char c = str[i];
   52                 int index = (c-‘a‘) %26;
   53                 if (cur->chars[index] == NULL){
   54                     return str;
   55                 }
   56                 if (cur->chars[index]->eow){
   57                     return str.substr(0,i+1);
   58                 }
   59                 cur = cur->chars[index];
   60             }
   61             return str;
   62         }
   63     };
   64 public:
   65     string replaceWords(vector<string>& dict, string sentence) {
   66         Trie trie;
   67         for (string str : dict){
   68             trie.insertWord(str);
   69         }
   70         
   71         string ans = "";
   72         int index = 0, beg = 0;   
   73         while ((index = sentence.find(‘ ‘,beg)) != string::npos){
   74             ans += trie.getShortestPrefix(sentence.substr(beg, index - beg)) + " ";
   75             beg = index + 1;
   76         }
   77         ans += trie.getShortestPrefix(sentence.substr(beg));
   78         return ans;
   79     }
   80 };