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给定一个仅包含数字 2-9
的字符串,返回所有它能表示的字母组合。答案可以按 任意顺序 返回。
给出数字到字母的映射如下(与电话按键相同)。注意 1 不对应任何字母。
示例 1:
输入:digits = "23" 输出:["ad","ae","af","bd","be","bf","cd","ce","cf"]
示例 2:
输入:digits = "" 输出:[]
示例 3:
输入:digits = "2" 输出:["a","b","c"]
提示:
0 <= digits.length <= 4
digits[i]
是范围['2', '9']
的一个数字。
class Solution {
public://先做一个映射unordered_map<int,string>map = {{0,""},{1,""},{2,"abc"},{3,"def"},{4,"ghi"},{5,"jkl"},{6,"mno"},{7,"pqrs"},{8,"tuv"},{9,"wxyz"},};vector<string>res;string path;void backtracking(string& digits,int index){//index 表示遍历第几个集合if(index == digits.size()){res.push_back(path);return;}int digit = digits[index] - '0';string text = map[digit];for(int i = 0;i < text.size();i++){path.push_back(text[i]);backtracking(digits,index+1); // index+1:表示下一层(第二个集合)里面选元素//为什么和i没关系呢?因为每一次都要从 新集合里从第一个选取。path.pop_back();}}vector<string> letterCombinations(string digits) {if(digits.size() == 0) return res;backtracking(digits,0);return res;}
};