[leetcode] 720. Longest Word in Dictionary

[leetcode] 720. Longest Word in Dictionary

Description

Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order.

If there is no answer, return the empty string. Example 1:

Input: words = ["w","wo","wor","worl", "world"]Output: "world"Explanation: The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".

Example 2:

Input: words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]Output: "apple"Explanation: Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".

Note:

All the strings in the input will only contain lowercase letters.The length of words will be in the range [1, 1000].The length of words[i] will be in the range [1, 30].

分析

题目的意思是：找出字符串数组中的一个词，该词由字符串数组中其他的字符串一次添加一个字符得到，求找出其中最长的那个。

直接对长度为1的单词调用递归函数,在递归中，还是先判断单词和mxLen关系来更新结果res，然后就是遍历所有字符，加到单词后面，如果在集合中存在，调用递归函数，结束后恢复状态。

代码

class Solution {public: string longestWord(vector& words) { unordered_set s(words.begin(),words.end()); string res; int mxlen=0; for(string word:words){ if(word.size()==1){ solve(s,word,mxlen,res); } } return res; } void solve(unordered_set s,string word,int& mxlen,string& res){ if(word.size()>mxlen){ mxlen=word.size(); res=word; }else if(word.size()==mxlen){ res=min(res,word); } for(char c='a';c<='z';c++){ word.push_back(c); if(s.count(word)){ solve(s,word,mxlen,res); } word.pop_back(); } }};

参考文献

​​[LeetCode] Longest Word in Dictionary 字典中的最长单词​​

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