LeetCode: Letter Combinations of a Phone Number [018]

【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"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

【题意】

给定一个一位数字[0-9]组合，确定其在拨号键盘上的所有可能的字母组合

【思路】

1. 创建键盘字符字典

2. 用深度优先搜索DFS找出所有组合

【代码】

class Solution {
public:
    map<int, vector<char> > dict;

    void dfs(vector<string>&result, string digits, int numIndex, string combination){
        //numIndex-当前正在匹配的数字的索引位
        //combination-当前已经生成的组合字符串
        if(numIndex<digits.size() && (digits[numIndex]==‘1‘|| digits[numIndex]==‘0‘))numIndex++;    //考虑1和0的情况
        if(numIndex==digits.size()){
            result.push_back(combination);
            return;
        }

        vector<char> chars = dict[digits[numIndex]];
        for(int i=0; i<chars.size(); i++){
            dfs(result, digits, numIndex+1, combination+chars[i]);
        }
    }

    vector<string> letterCombinations(string digits) {
        vector<string>result;
        dict[‘2‘].push_back(‘a‘);
        dict[‘2‘].push_back(‘b‘);
        dict[‘2‘].push_back(‘c‘);
        dict[‘3‘].push_back(‘d‘);
        dict[‘3‘].push_back(‘e‘);
        dict[‘3‘].push_back(‘f‘);
        dict[‘4‘].push_back(‘g‘);
        dict[‘4‘].push_back(‘h‘);
        dict[‘4‘].push_back(‘i‘);
        dict[‘5‘].push_back(‘j‘);
        dict[‘5‘].push_back(‘k‘);
        dict[‘5‘].push_back(‘l‘);
        dict[‘6‘].push_back(‘m‘);
        dict[‘6‘].push_back(‘n‘);
        dict[‘6‘].push_back(‘o‘);
        dict[‘7‘].push_back(‘p‘);
        dict[‘7‘].push_back(‘q‘);
        dict[‘7‘].push_back(‘r‘);
        dict[‘7‘].push_back(‘s‘);
        dict[‘8‘].push_back(‘t‘);
        dict[‘8‘].push_back(‘u‘);
        dict[‘8‘].push_back(‘v‘);
        dict[‘9‘].push_back(‘w‘);
        dict[‘9‘].push_back(‘x‘);
        dict[‘9‘].push_back(‘y‘);
        dict[‘9‘].push_back(‘z‘);

        dfs(result, digits, 0, "");
        return result;
    }
};

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