648. Replace Words

Problem statement

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"

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

Solution

Although this is the fourth problem of leetcode weekly contest 42, the solution is pretty straightforward.

The best solution should employ the trie tree, by now, I have no idea of that. Alternatively, I add all root words into a hash table and search by the key value.

The general idea:

  • Add all roots into hash table.
  • Separate the strings into single word.
  • Search from the start of each word to find if root exists in hash table.

Time complexity is O(n), n is the size of setence, space complexity is O(n + m), m is the size of root.

class Solution {
public:
    string replaceWords(vector<string>& dict, string sentence) {
        // add the dict into a set to search
        set<string> ht;
        for(auto dic : dict){
            ht.insert(dic);
        }
        // partition the sentences into words
        vector<string> words;
        for(int i = 0, j = 0; i < sentence.size() && j <= sentence.size(); j++){
            if(sentence[j] == ' ' || j == sentence.size()){
                words.push_back(sentence.substr(i, j - i));
                i = j + 1;
            }
        }
        // find root in dictionary
        for(int i = 0; i < words.size(); i++){
            for(int j = 0; j < words[i].size(); j++){
                if(ht.find(words[i].substr(0, j + 1)) != ht.end()){
                    words[i] = words[i].substr(0, j + 1);
                    break;
                }
            }
        }
        // build the whole sentence
        string ans = words[0];
        for(int i = 1; i < words.size(); i++){
            ans += " " + words[i];
        }
        return ans;
    }
};

 

posted @ 2017-07-26 04:37  蓝色地中海  阅读(546)  评论(0编辑  收藏  举报