LeetCode 17: Letter Combinations of a Phone Number

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"].

解题思路

思路一：使用递归：

class Solution {public:    vector<string> letterCombinations(string digits) {        int dlen = digits.size();        vector<string> result;        if (dlen == 0 || digits[0] == '1' || digits[0] == '0') {            return result;        }        // 递归        vector<string> temp = letterCombinations(digits.substr(1, dlen - 1));        int tempLen = temp.size();        // 当前数字对应的字符的起始位置        int i = digits[0] < '8' ? (digits[0] - '2') * 3 : (digits[0] - '2') * 3 + 1;         // 当前数字对应的字符个数        int len = (digits[0] == '9' || digits[0] == '7') ? 4 : 3;        if (tempLen == 0) {            if (dlen == 1) {                // 字符串中不包含 '0' 或 '1'                for (int k = 0; k < len; ++k) {                    string str(1, 'a' + i + k);                    result.push_back(str);                }            }        }        else {            for (int k = 0; k < len; ++k) {                string str(1, 'a' + i + k);                for (int j = 0; j < tempLen; ++j) {                    // 将当前字符拼接到每一个递归返回的字符串中                    result.push_back(str + temp[j]);                }            }        }        return result;    }};

思路二：使用循环：

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> ret;        if(digits == "")            return ret;        ret.push_back("");        vector<string> dict(10); //0~9        dict[2] = "abc";        dict[3] = "def";        dict[4] = "ghi";        dict[5] = "jkl";        dict[6] = "mno";        dict[7] = "pqrs";        dict[8] = "tuv";        dict[9] = "wxyz";        for(int i = 0; i < digits.size(); i ++)        {            int size = ret.size();            for(int j = 0; j < size; j ++)            {                string cur = ret[0];                ret.erase(ret.begin());                for(int k = 0; k < dict[digits[i]-'0'].size(); k ++)                {                    ret.push_back(cur + dict[digits[i]-'0'][k]);                }            }        }        return ret;    }};