[Leetcode]#17 Letter Combinations of a Phone Number

最新推荐文章于 2025-08-22 19:21:44 发布

jerseyma

最新推荐文章于 2025-08-22 19:21:44 发布

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CC 4.0 BY-SA版权

文章标签： leetcode string

本文链接：https://blog.youkuaiyun.com/jerseyma/article/details/48173217

本文介绍了一种解决电话号码字母组合问题的C++实现方法，通过构建映射表将数字对应的字母快速组合成所有可能的字符串。算法首先检查输入是否为空，接着初始化映射表并计算结果集合的大小，最后通过迭代填充结果集中的每个位置。

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//#17 Letter Combinations of a Phone Number
//4ms 6.39%
class Solution {
public:
    vector<string> letterCombinations(string digits) 
    {

       if(digits.empty())
        {
            vector<string> r;
            return r;
        }

        vector<string> v(10, "");
        v[0] = " ";
        v[1] = "1";
        v[2] = "abc";
        v[3] = "def";
        v[4] = "ghi";
        v[5] = "jkl";
        v[6] = "mno";
        v[7] = "pqrs";
        v[8] = "tuv";
        v[9] = "wxyz";

        int string_length = digits.size();
        //cout << "string_length == " << string_length << endl;
        int vector_length(1);
        for(int i=0; i<string_length; i++)
        {
            vector_length = vector_length * v[ int (digits[i] - '0') ].size();
        }
        //cout << "vector_length == " << vector_length << endl;

        string s(string_length, '0');
        vector<string> r(vector_length, s);
        //cout << "typical s == " << s << endl;

        int block_number(1);
        for(int i=0; i<string_length; i++)
        {
            int this_string_size = v[ int (digits[i] - '0') ].size();
            //cout << "this_string_size == " << this_string_size << endl;
            int segment_number = (vector_length/block_number) / this_string_size;
            //cout << "segment_number == " << segment_number << endl;
            for(int j=0; j<block_number; j++)
            {
                for(int k=0; k<this_string_size; k++)
                {
                    for(int l=0; l<segment_number; l++)
                    {               
                        r[j*(vector_length/block_number) + k*segment_number + l][i] = v[ int (digits[i] - '0') ][k];
                    }
                }
            }
            block_number = block_number * this_string_size;
            //cout << "finish of " << i << endl;
        }
        //cout << "ok?\n";
        return r;

    }
};