leetcode 648. Replace Words 公共前缀 + 暴力查询

朴灿烈づ我的快乐病毒、 2022-06-03 07:06 97阅读 0赞

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

本题题意很简单，就是寻找前缀然后做替换，这道题的本意使用公共前缀树来说，但是这里就不这么麻烦了，直接去做即可，为了加速搜索**，这里按照首字母做了26个分类，然后做了一个排序，**最后暴力搜索即可

**注意这里使用26个字幕做一个出事的分类，这个会加速计算**

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    bool cmp(string a, string b)
    {
        return a.length() < b.length();
    }
    
    class Solution 
    {
    public:
        string replaceWords(vector<string>& dictory, string sentence) 
        {
            sort(dictory.begin(), dictory.end(), cmp);
    
            vector<vector<string>> dict(26);
            for (string s : dictory)
                dict[s[0] - 'a'].push_back(s);
    
            stringstream ss(sentence);
            string res = "",one = "";
            while (ss >> one)
            {
                for (string t : dict[one[0] - 'a'])
                {
                    if (one.substr(0, t.length()) == t)
                    {
                        one = t;
                        break;
                    }
                }
                res += one + " ";
            }
            res.pop_back();
            return res;
        }
    };