[LeetCode] 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.

方法1：backtracking

和subset, combination问题一样的backtracking。唯一的区别是要先建立一个从数字到字母的转换表。这样每一层递归遍历当前digits[i]所对应的所有字母，并加入当前combination中传到下一层递归。

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

class Solution { public: vector<string> letterCombinations(string digits) { vector<string> lettComb; string dict[] = {" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; string comb(digits.size(),'\0'); findLettComb(digits, 0, dict, comb, lettComb); return lettComb; } void findLettComb(string &digits, int index, string dict[], string &comb, vector<string> &lettComb) { if(index==digits.size()) { lettComb.push_back(comb); return; } string lett= dict[digits[index] - '0']; for(int i=0; i<lett.size(); i++) { comb[index] = lett[i]; findLettComb(digits, index+1, dict, comb, lettComb); } } };

方法2：插入法

当已经获得digits[0:i-1]的所有letter combinations以后，加入digits[i]后新combinations为加入每个可能对应的字母加到之前的解集中。这里需要克隆多份之前的解集。

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

class Solution { public: vector<string> letterCombinations(string digits) { vector<string> lettComb; lettComb.push_back(""); string dict[] = {" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; for(int i=0; i<digits.size(); i++) { int n = lettComb.size(); string lett = dict[digits[i]-'0']; for(int j=0; j<n; j++) { for(int k=1; k<lett.size(); k++) { string comb = lettComb[j]; //clone lettComb[j] comb.push_back(lett[k]); lettComb.push_back(comb); } lettComb[j].push_back(lett[0]); } } return lettComb; } };