[LeetCode 题解]: LetterCombinations

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.

题解： 递归。首先将输入字符串解析成数字集合，记录字符串大小为Len。

生成的字符串长度依旧为Len。从首位开始判断该位可以填充的字符。

注意：本题首先需要对手机键盘的字符进行映射，作为辅助。

class Solution {
public:
    char vi[10][4];
    vector<string > out;
    int length;
    void getVi()
    {
        int i,j;
        for(i=0;i<=9;i++)
            for(j=0;j<4;j++)
                vi[i][j]=0;
        for(i=2;i<=7;i++)
        {
            for(j=0;j<3;j++)
            {
                vi[i][j]= 'a'+ 3*(i-2)+j;
            }
        }
        vi[0][0]=' ';
        vi[7][3]='s';
        for(i=8;i<=9;i++)
        {
            for(j=0;j<3;j++)
            {
                vi[i][j]= 'b' + 3*(i-2)+j;
            }
        }
        vi[9][3]='z';
    }
    
    void DFS(int len,string digits, string c)
    {
        if(len==length)
        {
            out.push_back(c);
            return ;
        }
        int count = digits[len]-'0';
        for(int i=0;i<4;i++)
        {
           if(vi[count][i]!=0)
               DFS(len+1,digits,c+vi[count][i]);
        }
    }
    vector<string> letterCombinations(string digits) {
        out.clear();
        length = digits.size();
        if(length==0) 
        {
            out.push_back("");
            return out;
        }
        if(digits.find('1')!=string::npos) return out;
        getVi();
        DFS(0,digits,"");
        return out;
    }
};

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