[leetcode]17. Letter Combinations of a Phone Number

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本文介绍了一种简单高效的迭代解决方案来生成电话号码的所有可能字母组合。通过遍历输入的数字字符串，利用预先定义好的数字到字母的映射表，逐步生成所有可能的字母组合。

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题目链接：https://leetcode.com/problems/letter-combinations-of-a-phone-number/#/description

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

class Solution{
public:
    vector<string> letterCombinations(string digits)
    {
        static const vector<string> buckets={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
        vector<string> res;
        if(digits.empty())
            return res;
        res.push_back("");
        for(int i=0;i<digits.size();i++)
        {
            int num=digits[i]-'0';
            if(num<0 || num>9) break;
            string buc=buckets[num];
            if(buc.empty()) continue;
            vector<string> tmp;
            for(int j=0;j<res.size();j++)
            {
                for(int k=0;k<buc.length();k++)
                {
                    tmp.push_back(res[j]+buc[k]);
                }
            }
            res.swap(tmp);
        }
        return res;
    }
};

解释：

Simple and efficient iterative solution.

Explanation with sample input "123"

Initial state:

result = {""}

Stage 1 for number "1":

result has {""}
candiate is "abc"
generate three strings "" + "a", ""+"b", ""+"c" and put into tmp,
tmp = {"a", "b","c"}
swap result and tmp (swap does not take memory copy)
Now result has {"a", "b", "c"}

Stage 2 for number "2":

result has {"a", "b", "c"}
candidate is "def"
generate nine strings and put into tmp,
"a" + "d", "a"+"e", "a"+"f",
"b" + "d", "b"+"e", "b"+"f",
"c" + "d", "c"+"e", "c"+"f"
so tmp has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
swap result and tmp
Now result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }

Stage 3 for number "3":

result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
candidate is "ghi"
generate 27 strings and put into tmp,
add "g" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
so, tmp has
{"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
"adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
"adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }
swap result and tmp
Now result has
{"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
"adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
"adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }

Finally, return result.