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English Version

题目描述

给出一个字符串数组 words 组成的一本英语词典。返回 words 中最长的一个单词,该单词是由 words 词典中其他单词逐步添加一个字母组成。

若其中有多个可行的答案,则返回答案中字典序最小的单词。若无答案,则返回空字符串。

 

示例 1:

输入:words = ["w","wo","wor","worl", "world"]
输出:"world"
解释: 单词"world"可由"w", "wo", "wor", 和 "worl"逐步添加一个字母组成。

示例 2:

输入:words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
输出:"apple"
解释:"apply" 和 "apple" 都能由词典中的单词组成。但是 "apple" 的字典序小于 "apply" 

 

提示:

  • 1 <= words.length <= 1000
  • 1 <= words[i].length <= 30
  • 所有输入的字符串 words[i] 都只包含小写字母。

解法

方法一:哈希表

用哈希表存放所有单词。遍历这些单词,找出长度最长且字典序最小的单词。

方法二:排序

优先返回符合条件、长度最长且字典序最小的单词,那么可以进行依照该规则,先对 words 进行排序,免去多个结果之间的比较。

Python3

class Solution:
    def longestWord(self, words: List[str]) -> str:
        cnt, ans = 0, ''
        s = set(words)
        for w in s:
            n = len(w)
            if all(w[:i] in s for i in range(1, n)):
                if cnt < n:
                    cnt, ans = n, w
                elif cnt == n and w < ans:
                    ans = w
        return ans

Java

class Solution {
    private Set<String> s;

    public String longestWord(String[] words) {
        s = new HashSet<>(Arrays.asList(words));
        int cnt = 0;
        String ans = "";
        for (String w : s) {
            int n = w.length();
            if (check(w)) {
                if (cnt < n) {
                    cnt = n;
                    ans = w;
                } else if (cnt == n && w.compareTo(ans) < 0) {
                    ans = w;
                }
            }
        }
        return ans;
    }

    private boolean check(String word) {
        for (int i = 1, n = word.length(); i < n; ++i) {
            if (!s.contains(word.substring(0, i))) {
                return false;
            }
        }
        return true;
    }
}

TypeScript

function longestWord(words: string[]): string {
    words.sort((a, b) => {
        const n = a.length;
        const m = b.length;
        if (n === m) {
            return a < b ? -1 : 1;
        }
        return m - n;
    });
    for (const word of words) {
        let isPass = true;
        for (let i = 1; i <= word.length; i++) {
            if (!words.includes(word.slice(0, i))) {
                isPass = false;
                break;
            }
        }
        if (isPass) {
            return word;
        }
    }
    return '';
}

Rust

impl Solution {
    pub fn longest_word(mut words: Vec<String>) -> String {
        words.sort_unstable_by(|a, b| (b.len(), a).cmp(&(a.len(), b)));
        for word in words.iter() {
            let mut is_pass = true;
            for i in 1..=word.len() {
                if !words.contains(&word[..i].to_string()) {
                    is_pass = false;
                    break;
                }
            }
            if is_pass {
                return word.clone();
            }
        }
        String::new()
    }
}

C++

class Solution {
public:
    string longestWord(vector<string>& words) {
        unordered_set<string> s(words.begin(), words.end());
        int cnt = 0;
        string ans = "";
        for (auto w : s) {
            int n = w.size();
            if (check(w, s)) {
                if (cnt < n) {
                    cnt = n;
                    ans = w;
                } else if (cnt == n && w < ans)
                    ans = w;
            }
        }
        return ans;
    }

    bool check(string& word, unordered_set<string>& s) {
        for (int i = 1, n = word.size(); i < n; ++i)
            if (!s.count(word.substr(0, i)))
                return false;
        return true;
    }
};

Go

func longestWord(words []string) string {
	s := make(map[string]bool)
	for _, w := range words {
		s[w] = true
	}
	cnt := 0
	ans := ""
	check := func(word string) bool {
		for i, n := 1, len(word); i < n; i++ {
			if !s[word[:i]] {
				return false
			}
		}
		return true
	}
	for w, _ := range s {
		n := len(w)
		if check(w) {
			if cnt < n {
				cnt, ans = n, w
			} else if cnt == n && w < ans {
				ans = w
			}
		}
	}
	return ans
}

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