Problem 320: Generalized Abbreviation

https://leetcode.com/problems/generalized-abbreviation/#/description

思路

• 对每个 character 来说，有两种情况：取或者不取。

• 对于每个 dfs，也有两种情况：count 增加，或者从 0 开始。

public class Solution {
    public List<String> generateAbbreviations(String word) {
        List<String> rst = new ArrayList<>();
        dfs(word, "", 0, 0, rst);

        return rst;
    }

    private void dfs(String word, String cur, int pos, int count, List<String> rst) {
        if (pos == word.length()) {
            if (count > 0) cur += count;
            rst.add(cur);
            return;
        } 

        dfs(word, cur, pos + 1, count + 1, rst);
        dfs(word, cur + (count > 0 ? count : "") + word.charAt(pos), pos + 1, 0, rst);
    }
}