720. Longest Word in Dictionary*

720. Longest Word in Dictionary*

​​https://leetcode.com/problems/longest-word-in-dictionary/​​

题目描述

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]​​.

C++ 实现 1

可以考虑先对字典进行排序, 规则是:

• 长度小的排前面
• 长度相等的, 但是字典序大的排前面

上述规则也可以倒过来, 比如(长度大的排前面, 同时长度相等的但字典序小的排前面).

然后从后向前遍历数组, 判断每个字符串是否能用其他字符串逐渐形成, 这一步使用哈希表完成. 从后向前, 当找到第一个满足条件的字符串, 直接返回即可.

class Solution {
private:
    struct Comp {
        bool operator()(const string &s1, const string &s2) {
            if (s1.size() < s2.size())
                return true;
            else if (s1.size() == s2.size())
                return (s1 > s2);
            return false;
        }
    };
public:
    string longestWord(vector<string>& words) {
        std::sort(words.begin(), words.end(), Comp());
        unordered_set<string> record(words.begin(), words.end());

        for (int i = words.size() - 1; i >= 0; --i) {
            bool found = true;
            for (int j = words[i].size() - 1; j > 0; j--)
                if (!record.count(words[i].substr(0, j))) {
                    found = false;
                    break;
                }
            if (found)
                return words[i];
        }
        return "";

    }
};

C++ 实现 2

速度更慢一些, 不排序, 而是对哈希表中的每一个字符串, 判断是否能用其他字符串逐渐形成, 同时还要判断是否满足题目中的条件.

class Solution {
public:
    string longestWord(vector<string>& words) {
        unordered_set<string> record(words.begin(), words.end());
        string res;
        for (auto &p : record) {
            if (p.size() >= res.size()) {
                bool contains = true;
                for (int i = 0; i < p.size(); ++ i) {
                    if (!record.count(p.substr(0, i + 1))) {
                        contains = false;
                        break;
                    }
                }
                if (contains) {
                    if (res.empty() || p.size() > res.size()) res = p;
                    else if (p.size() == res.size()) res = p < res ? p : res;
                } 
            }
        }
        return res;
    }
};