Letter Combinations of a Phone Number -- leetcode

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.

算法一： 用数制进位的思维做组合。

class Solution {public:    vector<string> letterCombinations(string digits) {        const char *map[] = {" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        int idx[sizeof(map)/sizeof(map[0])] = {0};        vector<string> result;        string item;        bool carry = false;        while (!carry) {                item.clear();                carry = true;                for (int i=0; i<digits.size(); ++i) {                        const char * p = &map[digits[i]-'0'][idx[i]];                        if (*p)                                item.append(1, *p);                        if (carry && *p) {                                ++idx[i];                                ++p;                                if (!*p)                                        idx[i] = 0;                                else                                        carry = false;                        }                }                result.push_back(item);        }        if (result.empty())                result.push_back(item);        return result;    }};

下面这一句，纯粹是为了满足leetcode的输入为空串("")时的test case：

        if (result.empty())                result.push_back(item);

看到这算法的执行时间，忍不住做了个截图，留个纪念。

算法二，字符逐步添加法

class Solution {public:    vector<string> letterCombinations(string digits) {        const char *map[] = {" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        vector<string> result;        result.push_back("");        for (int i=0; i<digits.size(); i++) {                const int idx = digits[i] - '0';                if (!*map[idx]) continue;                const size_t size = result.size();                for (int j=0; j<size; j++) {                        const char *p = map[idx];                        while (*++p) {                                result.push_back(result[j]);                                result[result.size()-1].append(1, *p);                        }                        result[j].append(1, *map[idx]);                }        }        return result;    }};

这个算法在leetcode上实际执行时间也为4 ms。

算法三，递归

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> result;        string item;        permutate(result, digits, 0, item);        return result;    }    void permutate(vector<string> &res, const string &digits, size_t idx, string &item) {        if (idx == digits.size())                return res.push_back(item);        static const char *map[] = {" ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        const char *p = map[digits[idx] - '0'];        while (*p) {                item.push_back(*p);;                permutate(res, digits, idx+1, item);                item.pop_back(); // c++ 11                ++p;        }    }};

这个算法执行时间也是4ms。看来这道题，怎么写都是4ms。

这三个算法，都包含了对数字1的处理，1对应着一个空字符串。

不过leetcode上test case并未包含1的情况。所以即使不对1作特殊处理，也能被AC。