leetcode 648. Replace Words 公共前缀

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

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;    }};