320. Generalized Abbreviation

最新推荐文章于 2022-04-25 09:17:09 发布

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#back tracking

此博客介绍了一个算法，用于生成给定单词的所有可能通用缩写形式，包括数字和字母组合，适用于不同长度的单词。

Problem

Write a function to generate the generalized abbreviations of a word.

Example:

Given word = "word", return the following list (order does not matter):

["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]

Solution

典型的 backtracking, 唯一要注意的是例子给的单词长度是各位数，要考虑单词很长的时候可以是两位数，这样就不能用 num + '0' 来表示当前的数字字符了。

class Solution {
    void helper(int begin, string& oneRst, vector<string>& rst, const string& word ){
        
        if(begin == word.size()) {
            rst.push_back(oneRst);
            return;
        }
        
        oneRst.push_back(word[begin]);
        helper(begin + 1, oneRst, rst, word);
        oneRst.pop_back();
        
        if( oneRst.empty() ||  oneRst.back() < '0' || oneRst.back() > '9' ){
            for( int numLen = 1; numLen <= word.size() - begin; numLen++) {
                string str = to_string(numLen);
                oneRst.append(str);
                helper(begin + numLen, oneRst, rst, word);
                oneRst.erase(oneRst.end() - str.size(), oneRst.end());
            }
        }
    }
    
public:
    vector<string> generateAbbreviations(string word) {
        string oneRst; 
        vector<string> rst;
        helper(0, oneRst, rst, word);
        return rst;
    }
};