Leetcode 17. Letter Combinations of a Phone Number

17. Letter Combinations of a Phone Number

• Total Accepted: 89600
• Total Submissions: 300111
• Difficulty: Medium

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.

思路：可以用回溯法，也可以直接先进先出地构造。这里注意c++中string数组的new方式。

代码：
FIFO构造：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> v;
        if(!digits.size()) return v;
        string *mapping=new string[10]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        int n=digits.size(),i;
        v.push_back("");
        for(i=0;i<n;i++){
            while(v[0].size()==i){
                string s=v[0];
                v.erase(v.begin());
                string temp=mapping[digits[i]-'0'];
                for(int j=0;j<temp.size();j++){
                    v.push_back(s+temp[j]);
                }
            }
        }
        return v;
    }
};

回溯法：

class Solution {
public:
    void letterCombinations(vector<string>& v,string digits,string s,string *mapping,int cur,int n){
        if(cur==n){
            v.push_back(s);
            return;
        }
        int i,num=digits[cur]-'0';
        string temp=mapping[num];
        for(i=0;i<temp.size();i++){
            letterCombinations(v,digits,s+temp[i],mapping,cur+1,n);
        }
    }
    vector<string> letterCombinations(string digits) {
        vector<string> v;
        int n=digits.size();
        if(!n) return v;
        string *mapping=new string[10]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        letterCombinations(v,digits,"",mapping,0,n);
        return v;
    }
};