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

1. Begin with the first character and then the number of characters abbreviated, which followed by the last character.
2. 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.
3. 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:

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

这道题让我们求单词的缩写形式，就是首尾字母加上中间字符的个数组成的新字符串，但是要求是不能有重复的缩写字符串，而且说明如果缩写字符串的长度并没有减小的话就保留原来的字符串，比如 god，缩写成 g1d 也没啥用，所以仍是 god。博主刚开始在研究题目中给的例子的时候有些疑惑，虽然知道 internal 和 interval 的缩写形式都是 i6l，会冲突，博主刚开始不明白的是，为什么不能一个是 i6l，一个是 in5l，这样不就不冲突了么，而题目中的缩写形式居然都是原字符串。后来才搞清楚题目原来是说只要有冲突的都不能用，而 internal 和 interval 是典型的死杠上的一对，i6l，in5l，int4l，inte3l，inter2l，统统冲突，而再往后的缩写长度就和原字符串一样了，所以二者就都保留了原样。理解了题意就好办了，由于每个单词的缩写形式中数字前面的字母个数不一定相同，所以用一个 pre 数组来记录每个单词缩写形式开头字母的长度，初始化都为1，然后先求出所有单词 pre 为1的缩写形式，再来进行冲突处理。遍历每一个缩写字符串，进行 while 循环，新建一个 HashSet，然后遍历其他所有字符串，所有发现冲突字符串，就把冲突字符串的坐标存入 HashSet 中，如果没有冲突，那么 HashSet 为空，直接 break 掉，如果有冲突，那么还要把当前遍历的位置i加入 HashSet 中，然后遍历 HashSet 中所有的位置，对其调用缩写函数，此时 pre 对应的值自增1，直到没有冲突存在为止，参见代码如下：

class Solution {
public:
    vector<string> wordsAbbreviation(vector<string>& dict) {
        int n = dict.size();
        vector<string> res(n);
        vector<int> pre(n, 1);
        for (int i = 0; i < n; ++i) {
            res[i] = abbreviate(dict[i], pre[i]);
        }
        for (int i = 0; i < n; ++i) {
            while (true) {
                unordered_set<int> st;
                for (int j = i + 1; j < n; ++j) {
                    if (res[j] == res[i]) st.insert(j);
                }
                if (st.empty()) break;
                st.insert(i);
                for (auto a : st) {
                    res[a] = abbreviate(dict[a], ++pre[a]);
                }
            }
        }
        return res;
    }
    string abbreviate(string s, int k) {
        return (k >= (int)s.size() - 2) ? s : s.substr(0, k) + to_string((int)s.size() - k - 1) + s.back();
    }
};

Github 同步地址：

#527

类似题目：

Minimum Unique Word Abbreviation

Valid Word Abbreviation

Generalized Abbreviation

Unique Word Abbreviation

参考资料：

https://leetcode.com/problems/word-abbreviation/

https://leetcode.com/problems/word-abbreviation/discuss/99792/HashMap-%2B-Trie-greater-O(nL)-solution

https://leetcode.com/problems/word-abbreviation/discuss/99782/Really-simple-and-straightforward-Java-solution

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