LeetCode每日一题(720. Longest Word in Dictionary)

Given an array of strings words representing an English Dictionary, return 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.

Note that the word should be built from left to right with each additional character being added to the end of a previous word.

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”.

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 30
• words[i] consists of lowercase English letters.

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本来以为是用 Trie 来做， 但是写到一半发现，对于任何一个 word， 我只需要检查除了最后一个字母之前的所有字母组成的 prefix 是不是在字典里存在就好了， 如果存在则证明当前 word 是合法的， 不存在就是非法的。提前对 words 按照 word 的长度和字母顺序进行排序可以让我们只遍历一次即可完成， 而且对比的时候只需要考虑长度， 不需要考虑字母顺序。

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use std::collections::HashSet;

impl Solution {
    pub fn longest_word(mut words: Vec<String>) -> String {
        words.sort_by(|w1, w2| {
            if w1.len() == w2.len() {
                return w1.cmp(&w2);
            }
            return w1.len().cmp(&w2.len());
        });
        let mut m = HashSet::new();
        m.insert("".to_owned());
        let mut ans = "".to_owned();
        for mut w in words {
            let last_char = w.pop().unwrap();
            if m.contains(&w) {
                w.push(last_char);
                if w.len() > ans.len() {
                    ans = w.clone();
                }
                m.insert(w);
            }
        }
        ans
    }
}
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