leetcode: Letter Combinations of a Phone Number

http://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/

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：

穷举法加回溯。用一个二维数组保存每个数字对应可能出现的字母，当每一层穷举完后回溯到上一层，再次穷举下个字符可能出现的组合。

static char keys[8][4] = {{'a', 'b', 'c'},       // 2
                         {'d', 'e', 'f'},       // 3
                         {'g', 'h', 'i'},       // 4
                         {'j', 'k', 'l'},       // 5
                         {'m', 'n', 'o'},       // 6
                         {'p', 'q', 'r', 's'},  // 7
                         {'t', 'u', 'v'},       // 8
                         {'w', 'x', 'y', 'z'}}; // 9
static int sizes[] = {3, 3, 3, 3, 3, 4, 3, 4}; 

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> result;
        int levels = digits.size(), current_level = 0;
        vector<int> indexes(levels, 0);
        
        if (0 == levels) {
            return vector<string>(1, "");
        }
        
        while (true) {
            int ch_index = digits[current_level] - '0' - 2;
            
            if (sizes[ch_index] == indexes[current_level]) {
                if (0 == current_level) {
                    break;
                }
                
                indexes[current_level] = 0;
                --current_level;
                ++indexes[current_level];
                continue;
            }
            else {
                if (current_level < (levels - 1)) {
                    ++current_level;
                }
                else {
                    string s;
                    int i;
                    
                    for (i = 0; i < current_level; ++i) {
                        s.push_back(keys[digits[i] - '0' - 2][indexes[i]]);
                    }
                    s.push_back('?');
                    
                    for (i = 0; i < sizes[ch_index]; ++i) {
                        s[current_level] = keys[ch_index][i];
                        result.push_back(s);
                    }
                    
                    indexes[current_level] = sizes[ch_index];
                }
            }
        }
        
        return result;
    }
};