17. Letter Combinations of a Phone Number

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分类专栏： leetcode

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本文介绍了一种使用递归算法生成电话号码对应的所有可能字母组合的方法。通过建立数字到字母的映射表，并利用递归函数进行遍历，可以有效地找出输入数字字符串的所有可能组合。

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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”].

模拟，用递归就ok啦

class Solution {
public:
    char change(char c, int a)
    {
        char n[300][300];
        n['2'][0] = '3',n['2'][1] = 'a',n['2'][2] = 'b',n['2'][3] = 'c';
        n['3'][0] = '3',n['3'][1] = 'd',n['3'][2] = 'e',n['3'][3] = 'f';
        n['4'][0] = '3',n['4'][1] = 'g',n['4'][2] = 'h',n['4'][3] = 'i';
        n['5'][0] = '3',n['5'][1] = 'j',n['5'][2] = 'k',n['5'][3] = 'l';
        n['6'][0] = '3',n['6'][1] = 'm',n['6'][2] = 'n',n['6'][3] = 'o';
        n['7'][0] = '4',n['7'][1] = 'p',n['7'][2] = 'q',n['7'][3] = 'r',n['7'][4] = 's';
        n['8'][0] = '3',n['8'][1] = 't',n['8'][2] = 'u',n['8'][3] = 'v';
        n['9'][0] = '4',n['9'][1] = 'w',n['9'][2] = 'x',n['9'][3] = 'y',n['9'][4] = 'z';
        n['#'][0] = '2',n['#'][1] = 'h',n['#'][2] = 'X'; 

        return n[c][a];
    }
    vector<string> s;
    int len;
    void getit(string d, int n, string ss)
    {
        int y = change(d[n - 1], 0) - '0';
        if(n == len)
        {
            for(int i = 1; i <= y; ++ i)
                s.push_back(ss + change(d[n - 1], i));
            return ;
        }
        for(int i = 1; i <= y; ++ i)
            getit(d, n + 1, ss + change(d[n - 1], i));
        return ;
    }
    vector<string> letterCombinations(string digits) {
        len = digits.size();
        if(len < 1) return s;
        getit(digits, 1, "");
        return s;
    }
};