以后会更新对应的栈操作实现。 首先放出原题:Letter Combinations of a Phone Number Given a string containing digits from2-9inclusive, 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. Note th...
* 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"]. */ public class LetterCombinationsofaPhoneNumber { private char[]...
['w','x','y','z']}# 回溯defback_trace(combination,next_digits):# 限定条件# 确定组合iflen(next_digits)==0:res.append(combination)else:# 遍历匹配数字的字母foreleminphone[next_digits[0]]:# 将字母组合起来# 继续下一个字母back_trace(combination+elem,next_digits[1:])res=[]ifdigits:back...
class LetterCombinationsPhoneNumber {func letterCombinations(_ digits: String) -> [String] {guard digits.count> 0else{return[String]()}var combinations = [String](), combination =""let numberToStr = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]dfs(&combinations, &...
[10]{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};voidtrack_back(vector<string>&result,string combination,string next){if(next.size()==0)result.push_back(combination);elsefor(autoletter:table[next[0]-'0'])track_back(result,combination+letter,next.substr(1));}...
{'9',"wxyz"}};string combination;backtrack(combinations,phoneMap,digits,0,combination);returncombinations;}voidbacktrack(vector<string>&combinations,constunordered_map<char,string>&phoneMap,conststring&digits,intindex,string&combination){if(index==digits.length()){combinations.push_back(combination);...
(list_temp)*len(last))]#initialthelastcombinationifcur_y==0:last=list_templast_new=last.copy()else:num=0#combinationforiinrange(len(last)):forjinrange(len(list_temp)):last_new[num]=last[i]+list_temp[j]num=num+1cur_y=cur_y+1last=last_new.copy()dyn_loop(list2,cur_y,last)...
大家少用全局变量，力扣的检测机制似乎是只有一次程序运行，但你的代码执行多变，然后一次性输入所有测试用例，用全局变量可能会导致两次测试用例相互干扰，如果一定要使用全局变量请务必保证每次重新执行你的代码段时你设置的全局变量会被初始化（我就用了个全局变量，然后点运行时是正确的，提交时是错误的）
Combination Sum/组合总和 给定一个无重复元素的数组 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target 的组合。candidates 中的数字可以无限制重复被选取。 Combination Sum II/组合总和 II 给定一个数组 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target ...
forletterinphone[next_digits[0]]: # append the current letter to the combination # and proceed to the next digits backtrack(combination+letter,next_digits[1:]) output=[] ifdigits: backtrack("",digits) returnoutput 1. 2. 3. 4.