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：

 1 class Solution {
 2 public:
 3     vector<string> letterCombinations(string digits) {
 4         string digMap[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
 5         vector<string> v;
 6         if (digits.size() == 0) {
 7             return v;
 8         }
 9         queue<string> q;
10         for (int i = 0; i < digMap[digits[0] - ‘0‘].size(); ++i) {
11             q.push(string(1,digMap[digits[0] - ‘0‘][i] ));
12         }
13         for (int i = 1; i < digits.size(); ++i) {
14             int l = q.size();
15             for (int j = 0; j < l; ++j) {
16                 string temp1 = q.front();
17                 q.pop();
18                 for (int k = 0; k < digMap[digits[i] - ‘0‘].size(); ++k) {
19                     string temp2 = temp1 + digMap[digits[i] - ‘0‘][k];
20                     q.push(temp2);
21                 }
22             }
23         }
24         while (!q.empty()) {
25             v.push_back(q.front());
26             q.pop();
27         }
28         return v;
29     }
30 };

回溯法：

 1 class Solution {
 2 private:
 3     string digMap[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
 4     void dfs(vector<string>& v, string& tempStr, int index, string& digits) {
 5         if (index == digits.size()) {
 6             v.push_back(tempStr);
 7             return;
 8         }
 9         for (int i = 0; i < digMap[digits[index] - ‘0‘].size(); ++i) {
10             //tempStr += digMap[digits[index] - ‘0‘][i];
11             tempStr.push_back(digMap[digits[index] - ‘0‘][i]); //string的push_back和pop_back以前没用过
12             dfs(v,tempStr,index + 1,digits);
13             tempStr.pop_back();
14         }
15     }
16 public:
17     vector<string> letterCombinations(string digits) {
18         vector<string> result;
19         if (digits.size() == 0) {
20             return result;
21         }
22         string tempStr;
23         dfs(result,tempStr,0,digits);
24         return result;
25     }
26 };