17. 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.

分析

方法一

每个字母都会独自展开一个分支，就是排列组合问题。DFS搜索

    "425"

      |

g     h     i         4

              /|   /|   /|         

             a b c a b c a b c       7

                                     .

                      .

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class Solution { public:     vector<string> alphabets = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};     vector<string> letterCombinations(string digits) {         int len = digits.size();         vector<string> result;         if( len == 0) return result;         dfs(digits, 0, "", result);                   return result;               }     void dfs(string digits, int pos, string path, vector<string> & result){         if(digits.size() == pos){             result.push_back(path);             return;         }         for(auto d: alphabets[digits[pos] - '0']){             dfs(digits, pos + 1, path + d, result);         }     } };

方法二

转自：http://www.tuicool.com/articles/m6BFNfR

可以先将0...1的所有排列找到-->{"a", "b", "c"}

再进一步将0...2的所有排列找到-->{"ad", "ae","af", "bd", "be", "bf", "cd", "ce", "cf"}

如此循环...直到字符串末尾。实现如下：

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class Solution { public:     vector<string> charmap = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};     vector<string> letterCombinations(string digits) {         vector<string> res;         if(digits.size() == 0) return res;         res.push_back("");         for (int i = 0; i < digits.size(); i++)         {             vector<string> tempres;             string chars = charmap[digits[i] - '0'];             for (int c = 0; c < chars.size();c++)                 for (int j = 0; j < res.size();j++)                     tempres.push_back(res[j]+chars[c]);             res = tempres; //重新赋值结果vector数组         }         return res;     } };

来自为知笔记(Wiz)