472. Concatenated Words

最新推荐文章于 2020-12-14 21:35:40 发布

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最新推荐文章于 2020-12-14 21:35:40 发布

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文章标签： string recursion dynamic programming

本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/90285905

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博客介绍了一种处理字符串的方法。给定数组元素数量不超10000，元素长度总和不超600000，输入仅含小写字母。采用递归方法，类似word break，先确认左半部分可查，再递归拆解右半部分，查找前删掉单词本身。

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472. Concatenated Words

方法1: recursion

Given a list of words (without duplicates), please write a program that returns all concatenated words in the given list of words.
A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.

Example:

Input: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]

Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]

Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; 
 "dogcatsdog" can be concatenated by "dog", "cats" and "dog"; 
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

Note:

The number of elements of the given array will not exceed 10,000
The length sum of elements in the given array will not exceed 600,000.
All the input string will only include lower case letters.
The returned elements order does not matter.

方法1: recursion

思路：

和word break一样，先取左边确认可查，然后递归拆解右半部分。这里需要在每次查找前把单词本身删掉，也不需要查找单词自身（左右空子串的情况都不用考虑），

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        unordered_set<string> dict(words.begin(), words.end());
        vector<string> result;
        for (auto word: words) {
            dict.erase(dict.find(word));
            if (findAllHelper(word, dict)) {
                result.push_back(word);
            }
            dict.insert(word);
        }
        return result;
    }
    
    bool findAllHelper(string s, unordered_set<string> & dict) {
        if (dict.count(s)) return true;
        
        for (int i = 1; i < s.size(); i++) {
            string left = s.substr(0, i);
            string right = s.substr(i);
            if (dict.count(left) && findAllHelper(right, dict)) {
                return true;
            }
        }
        return false;
    }
};