leetcode 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].

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 "";
    }
};