527. Word Abbreviation

方法1:
- Complexity

Given an array of n distinct non-empty strings, you need to generate minimal possible abbreviations for every word following rules below.

Begin with the first character and then the number of characters abbreviated, which followed by the last character.
If there are any conflict, that is more than one words share the same abbreviation, a longer prefix is used instead of only the first character until making the map from word to abbreviation become unique. In other words, a final abbreviation cannot map to more than one original words.
If the abbreviation doesn’t make the word shorter, then keep it as original.

Example:
Input: [“like”, “god”, “internal”, “me”, “internet”, “interval”, “intension”, “face”, “intrusion”]

Output: [“l2e”,“god”,“internal”,“me”,“i6t”,“interval”,“inte4n”,“f2e”,“intr4n”]

Note:

Both n and the length of each word will not exceed 400.
The length of each word is greater than 1.
3l.The words consist of lowercase English letters only.
The return answers should be in the same order as the original array.

方法1:

grandyang: https://www.cnblogs.com/grandyang/p/6818742.html

思路：

所谓的greedy，我们先尝试用一个prefix来压缩所有单词，然后开始检查是否有重复，如果有就进一步解压缩，直到没有collision的一颗。那么怎么实现呢，首先要用一个新的vector<\string>来记录当前压缩的每个单词结果。对于每一个压缩，开始向后面所有的单词发起遍历，如果发现一样的缩写，暂时将所有一样的index存起来，在遍历结束后对所有对应单词进一步压缩。那么又会需要一个vector<\int>来对每一个单词记录迄今为止压缩的prefix位置，每一次进一步压缩要所有重复一起++。

Complexity

class Solution {public:vector<string> wordsAbbreviation(vector<string>& dict) {int n = dict.size();vector<int> prefix(n, 1);vector<string> res;for (string word: dict) {if (word.size() <= 3) res.push_back(word);else {res.push_back(*word.begin() + to_string(word.size() - 2) + *word.rbegin());}}for (int i = 0; i < n; i++) {while (true) {vector<int> dup;for (int j = i + 1; j < n; j++) {if (res[i] == res[j]) dup.push_back(j);}if (dup.empty()) break;dup.push_back(i);for (int k: dup) res[k] = abbrev(dict[k], ++prefix[k]);}}return res;}string abbrev(string & word, int pre) {return (pre >= word.size() - 2) ? word : (word.substr(0, pre) + to_string(word.size() - pre - 1) + word.back());}
};

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527. Word Abbreviation

527. Word Abbreviation

方法1:

Complexity

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