leetcode 720. Longest Word in Dictionary

最新推荐文章于 2025-08-22 01:16:35 发布

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最新推荐文章于 2025-08-22 01:16:35 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/pk28/p/8462085.html

本文介绍了一种算法，用于从给定的单词列表中找出能通过其他单词逐步构建的最长单词。若存在多个符合条件的单词，则返回字典序最小的一个。

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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].

题目大意：就是找一个最长的字符串，并且这个字符串是由其他的字符串一步一步转换得到的。
思路：先排序，按照长度再按照字母顺序。然后从后往前找，找到的第一个符合要求的就是答案。
代码如下：

class Solution {
public:
    string longestWord(vector<string>& words) {
        sort(words.begin(), words.end(),[&](string a, string b){
            if (a.size() == b.size()) {
                return a > b;
            }
            return a.size() < b.size();
        });
        int n = words.size();
        unordered_map<string, int> mp;
        for (int i = 0; i < n; ++i) {
            mp[words[i]]++;
        }
        for (int i = n-1; i >=0; --i) {
            mp[words[i]]--;
            int mark = 0;
            int m = words[i].size();
            for (int j = m-1; j >= 1; --j) {
                string s = words[i].substr(0, j);
                if (!mp[s]) {
                    mark = 1;
                    break;
                }
            }
            if (mark == 0) return words[i];
        }
        return "";
    }
};

转载于:https://www.cnblogs.com/pk28/p/8462085.html