17-Letter Combinations of a Phone Number

题目

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”].

分析

这个也是用到DFS，这是自己写的比较快的一个程序，因为之前被虐的够惨，像Permutation 系列，N-Queens 系列都是这个套路。

递归解决问题，不过我写的函数总是参数特别多。

值得一提的是，在这个函数里第一次用到string 的push_back() 和 pop_back()函数，原本以为只有vector 才有，是我孤陋寡闻了·······

实现

class Solution {
public:
vector letterCombinations(string digits) {

//存放结果序列    vector<string> result;    if (digits.length() == 0)        return result;//索引    string dict[] = {"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};//中间结果，也就是一个string    string subResult="";    letterCombinationsDFS(result,dict,digits,subResult,0);    return result;}void letterCombinationsDFS(vector<string>& result,string *dict,string& digits,string& subResult,int num){    if (num == digits.length())    {        result.push_back(subResult);        return;    }    else    {        string tmp = dict[digits[num] - '2'];        for (int i = 0; i < tmp.length(); i++)        {        //向string 的end 添加一个字符            subResult.push_back(tmp[i]);            letterCombinationsDFS(result, dict, digits, subResult, num + 1);        //从string的末端删除一个字符            subResult.pop_back();        }    }}

};