Leetcode 17. Letter Combinations of a Phone Number(python+cpp)

题目

解析

这题实际上是一个无重复combination的模板题，只需要对1做一下特殊处理即可
关于combination和permutation的总结，看这里：

代码：

class Solution:def letterCombinations(self, digits: str) -> List[str]:def backtracking(comb,level):if level==len(digits):ans.append(comb[:])returnfor c in d[digits[level]]:#comb.appned(c)backtracking(comb+c,level+1)#comb.pop()d = {
    '2':['a','b','c'],'3':['d','e','f'],'4':['g','h','i'],'5':['j','k','l'],'6':['m','n','o'],'7':['p','q','r','s'],'8':['t','u','v'],'9':['w','x','y','z']}if not digits:return []ans = []backtracking('',0)return ans

C++版本：
注意一下unordered_map的直接初始化

class Solution {
    
public:vector<string> letterCombinations(string digits) {
    if(digits.size() < 1) return {
    };unordered_map<char,string> mapping{
    {
    '2',"abc"},{
    '3',"def"},{
    '4',"ghi"},{
    '5',"jkl"},{
    '6',"mno"},{
    '7',"pqrs"},{
    '8',"tuv"},{
    '9',"wxyz"}};vector<string> ans;backtracking(mapping,ans,digits,0,"");return ans;}void backtracking(unordered_map<char,string>& mapping, vector<string>& ans, string digits, int level, string curr){
    if(level == digits.size()){
    ans.push_back(curr);return;}// for(int i=level;i<digits.size();i++){
    for(auto c : mapping[digits[level]]){
    curr += c;backtracking(mapping,ans,digits,level+1,curr);curr.pop_back();}// }}
};