LeetCode17 Letter Combinations of a Phone Number

题意：

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.(Medium)

分析：

很好的练习搜索的题目，可以用BFS和回溯法（DFS）两种方式来做一下练习。

自己写的时候没太想明白回溯的写法（参数，返回值等等），写的是BFS的解法，DFS参考的讨论区。

代码1：

 class Solution {
 public:
     vector<string> letterCombinations(string digits) {
         string digMap[] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
         vector<string> v;
         if (digits.size() == ) {
             return v;
         }
         queue<string> q;
         for (int i = ; i < digMap[digits[] - ''].size(); ++i) {
             q.push(string(,digMap[digits[] - ''][i] ));
         }
         for (int i = ; i < digits.size(); ++i) {
             int l = q.size();
             for (int j = ; j < l; ++j) {
                 string temp1 = q.front();
                 q.pop();
                 for (int k = ; k < digMap[digits[i] - ''].size(); ++k) {
                     string temp2 = temp1 + digMap[digits[i] - ''][k];
                     q.push(temp2);
                 }
             }
         }
         while (!q.empty()) {
             v.push_back(q.front());
             q.pop();
         }
         return v;
     }
 };

回溯法：

 class Solution {
 private:
     string digMap[] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
     void dfs(vector<string>& v, string& tempStr, int index, string& digits) {
         if (index == digits.size()) {
             v.push_back(tempStr);
             return;
         }
         for (int i = ; i < digMap[digits[index] - ''].size(); ++i) {
             //tempStr += digMap[digits[index] - '0'][i];
             tempStr.push_back(digMap[digits[index] - ''][i]); //string的push_back和pop_back以前没用过
             dfs(v,tempStr,index + ,digits);
             tempStr.pop_back();
         }
     }
 public:
     vector<string> letterCombinations(string digits) {
         vector<string> result;
         if (digits.size() == ) {
             return result;
         }
         string tempStr;
         dfs(result,tempStr,,digits);
         return result;
     }
 };