Problem: 17. 电话号码的字母组合

题目描述

思路及解法

1.将电话号码和对应的数组存入数组中创建映射关系；
2.编写，并调用回溯函数，当决策阶段等于digits的长度时，将当前的决策路径添加到结果集合中，否则从digits字符串的第一个字符开始回溯调用！我们在回溯时利用一个变量记录决策阶段，每次取出对应决策阶段可能的字母，并在此阶段上回溯

复杂度

时间复杂度:

最坏时间复杂度：$O(3^m\times 4^n)$,其中m表示选择对应只有三个字母的数字的个数，n表示对应四个字母的个数

空间复杂度:

$O(n+m)$

Code

class Solution {//Recode the resultvector<string> res;//Recode the decision pathvector<char> track;
public:/*** Get the all possible combinations** @param digits The combination given by topic* @return vector<string>*/vector<string> letterCombinations(string digits) {if (digits.size() == 0) {return {};}//Create the reflectionvector<string> mappings(10);mappings[2] = "abc";mappings[3] = "def";mappings[4] = "ghi";mappings[5] = "jkl";mappings[6] = "mno";mappings[7] = "pqrs";mappings[8] = "tuv";mappings[9] = "wxyz";backtrack(mappings, digits, 0, track);return res;}/*** Use backtracking to find all possible combinations** @param mappings Mapping between letters and numbers* @param digits Combination of numbers given by topic* @param stage Decision stage* @param track Decision path*/void backtrack(vector<string> &mappings, string &digits, int stage, vector<char> &track) {//End conditionif (stage == digits.length()) {res.push_back(string(track.begin(), track.end()));return;}string mapping = mappings[digits[stage] - '0'];for (int i = 0; i < mapping.length(); ++i) {track.push_back(mapping[i]);backtrack(mappings, digits, stage + 1, track);track.pop_back();}}
};