720. Longest Word in Dictionary

 

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].

 

这道题给了我们一个字典,是个字符串数组,然后问我们从单个字符开始拼,最长能组成啥单词,注意中间生成的字符串也要在字典中存在,而且当组成的单词长度相等时,返回字母顺序小的那个。好,看到这么多前缀一样多字符串,是不是很容易想到用前缀树来做,其实我们并不需要真正的建立前缀树结点,可以借鉴查找的思想来做。那么为了快速的查找某个单词是否在字典中存在,我们将所有单词放到哈希集合中,在查找的时候,可以采用BFS或者DFS都行。先来看BFS的做法,使用一个queue来辅助,我们先把所有长度为1的单词找出排入queue中,当作种子选手,然后我们进行循环,每次从队首取一个元素出来,如果其长度大于我们维护的最大值mxLen,则更新mxLen和结果res,如果正好相等,也要更新结果res,取字母顺序小的那个。然后我们试着增加长度,做法就是遍历26个字母,将每个字母都加到单词后面,然后看是否在字典中存在,存在的话,就加入queue中等待下一次遍历,完了以后记得要恢复状态,参见代码如下:

 

解法一:

class Solution {
public:
    string longestWord(vector<string>& words) {
        string res = "";
        int mxLen = 0;
        unordered_set<string> s(words.begin(), words.end());
        queue<string> q;
        for (string word : words) {
            if (word.size() == 1) q.push(word);
        }
        while (!q.empty()) {
            string t = q.front(); q.pop();
            if (t.size() > mxLen) {
                mxLen = t.size();
                res = t;
            } else if (t.size() == mxLen) {
                res = min(res, t);
            }
            for (char c = 'a'; c <= 'z'; ++c) {
                t.push_back(c);
                if (s.count(t)) q.push(t);
                t.pop_back();
            }
        }
        return res;
    }
};

 

下面来看递归的解法,前面都一样,不同在于直接对长度为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,参见代码如下:

 

解法二:

class Solution {
public:
    string longestWord(vector<string>& words) {
        string res = "";
        int mxLen = 0;
        unordered_set<string> s(words.begin(), words.end());for (string word : words) {
            if (word.size() == 1) helper(s, word, mxLen, res);
        }
        return res;
    }
    void helper(unordered_set<string>& 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)) helper(s, word, mxLen, res);
            word.pop_back();
        }
    }
};

 

下面这种解法是论坛上的高分解法,其实我们只要给数组排个序,就可以使用贪婪算法来做了,并不需要什么DFS或BFS这么复杂。首先建立一个空的哈希set,然后我们直接遍历排序后的字典,对于当前的单词,如果当前单词长度为1,或者该单词去掉最后一个字母后在集合中存在,这也不难理解,长度为1,说明是起始单词,不需要的多余的判断,否则的话就要判断其去掉最后一个字母后的单词是否在集合中存在,存在的话,才说明其中间单词都存在,因为此时是从短单词向长单词遍历,只要符合要求的才会加入集合,所以一旦其去掉尾字母的单词存在的话,那么其之前所有的中间情况都会在集合中存在。我们更新结果res时,要判断当前单词长度是否大于结果res的长度,因为排序过后,默认先更新的字母顺序小的单词,所有只有当当前单词长度大,才更新结果res,之后别忘了把当前单词加入集合中,参见代码如下:

 

解法三:

class Solution {
public:
    string longestWord(vector<string>& words) {
        string res = "";
        unordered_set<string> s;
        sort(words.begin(), words.end());
        for (string word : words) {
            if (word.size() == 1 || s.count(word.substr(0, word.size() - 1))) {
                res = (word.size() > res.size()) ? word : res;
                s.insert(word);
            }
        }
        return res;
    }
};

 

类似题目:

Longest Word in Dictionary through Deleting

Implement Magic Dictionary

 

参考资料:

https://discuss.leetcode.com/topic/109643/java-c-clean-code

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation