Letter Combinations of a Phone Number--LeetCode

1.题目

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.

2.题意

给定一个数字字符串, 返回该数字可以表示的所有可能的字母组合
下面给出了数字到字母的映射 (就像在电话按钮上一样)
string dict[] = {“abc”, “def”, “ghi”, “jkl”, “mno”, “pqrs”, “tuv”, “wxyz”};
注意:
虽然上面的答案是字典的, 但你的答案可能是你想要的任何顺序

3.分析

排列组合问题
1）递归版
建立字典保存每个数字所代表的字符串
用一个变量level，记录当前生成的字符串的字符个数
递归出口为level == digits.size()
注意letterCombinationsDFS(result, dict, digits, level + 1, temp);
不要写成letterCombinationsDFS(result, dict, digits, level + 1, str);
level == digits.size()之后result.push_back(temp);再return;不要写成return result;

2）迭代版
依次读取数字
将数字可以代表的字符加入当前结果
然后进入下一次迭代
注意提前result.push_back(“”);
string temp = result.front();不要写成string temp = result.begin();
不要忘记return reuslt;

4.代码

1）递归版

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> result;        if(digits.size() == 0)            return result;        string dict[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        letterCombinationsDFS(result, dict, digits, 0, "");        return result;    }    void letterCombinationsDFS(vector<string> &result, string dict[],        string digits, int level, string temp)    {        if(level == digits.size())        {            result.push_back(temp);            return;        }        string str = dict[digits[level] - '2'];        for(int i = 0; i < str.size(); ++i)        {            temp.push_back(str[i]);            letterCombinationsDFS(result, dict, digits, level + 1, temp);            temp.pop_back();        }        return;    }};  

2）迭代版

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> result;        if(digits.size() == 0)            return result;        string dict[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        result.push_back("");        for(int i = 0; i < digits.size(); ++i)        {            int n = result.size();            string str = dict[digits[i] - '2'];            for(int j = 0; j < n; ++j)            {                string temp = result.front();                result.erase(result.begin());                for(int k = 0; k < str.size(); ++k)                {                    result.push_back(temp + str[k]);                }            }        }        return result;    }};