前言我们社区会继续解决谷一（Netflix成长黑客，《iOS 面试之道》作者，ACE职业健身教练。微博：@古谣道长[1]）的Swift算法题整理成文字版，方便大家学习阅读。到目前为止，我们已经更新了16期的LeetCode算法。我们会保持更新时间和进度（周一、周三、周五上午9:00发布）。每期内容不多。希望大家在上班的路上读一读，积累久了会有很大的提升。不积步，无以至万里；不积小流，则不成江海。Swift社区将陪伴您一路前行。如果您有建议和意见，请在文末留言，我们将尽力满足您的需求。难度等级：中等1.描述给定一个只包含数字2-9的字符串，返回它能代表的所有字母组合。可以按任何顺序返回答案。数字到字母的映射如下（与电话键相同）。请注意，1不对应任何字母。2.示例示例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"]约束：0<=digits.length<=4digits[i]是['2','9']范围内的数字3。答案classLetterCombinationsPhoneNumber{funcletterCombinations(_digits:String)->[String]{guarddigits.count>0else{return[String]()}varcombinations=[String](),combination=""letnumberToStr=["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]dfs(&combinations,&combination,numberToStr,digits,0)returncombinations}privatefuncdfs(_combinations:inout[String],_combination:inoutString,_numberToStr:[String],_digits:String,_index:Int){ifcombination.count==digits.count{combinations.append(combination)return}letcurrentStr=fetchCurrentStr(from:digits,at:index,numberToStr)forcharincurrentStr{combination.append(char)dfs(&combinations,&combination,numberToStr,digits,index+1)combination.removeLast()}}privatefuncfetchCurrentStr(从数字:String,atindex:Int,_numberToStr:[String])->String{guardindex>=0&&index
