Letter Combinations of a Phone Number

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"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

 

枚举所有情况。

对于每一个输入数字，对于已有的排列中每一个字符串，分别加入该数字所代表的每一个字符。

所有是三重for循环。

举例：

初始化排列{""}

1、输入2，代表"abc"

已有排列中只有字符串""，所以得到{"a","b","c"}

2、输入3，代表"def"

(1)对于排列中的首元素"a"，删除"a"，并分别加入'd','e','f'，得到{"b","c","ad","ae","af"}

(2)对于排列中的首元素"b"，删除"b"，并分别加入'd','e','f'，得到{"c","ad","ae","af","bd","be","bf"}

(2)对于排列中的首元素"c"，删除"c"，并分别加入'd','e','f'，得到{"ad","ae","af","bd","be","bf","cd","ce","cf"}

注意

（1）每次添加新字母时，应该先取出现有ret当前的size()，而不是每次都在循环中调用ret.size()，因为ret.size()是不断增长的。

（2）删除vector首元素代码为：

ret.erase(ret.begin());

 

 

class Solution {public:    vector
     
       letterCombinations(string digits) {        vector
      
        ret;        if(digits == "")            return ret;                    ret.push_back("");                vector
       
         dict(10); //0~9        dict[2] = "abc";        dict[3] = "def";        dict[4] = "ghi";        dict[5] = "jkl";        dict[6] = "mno";        dict[7] = "pqrs";        dict[8] = "tuv";        dict[9] = "wxyz";                for(int i = 0; i < digits.size(); i ++)        {            int size = ret.size();            for(int j = 0; j < size; j ++)            {                string cur = ret[0];                ret.erase(ret.begin());                for(int k = 0; k < dict[digits[i]-'0'].size(); k ++)                {                    ret.push_back(cur + dict[digits[i]-'0'][k]);                }            }        }        return ret;    }};