这样的话,显然array是一个比较好的选择。根据测试, 数字1和0都是对应空 “”。 2.因为是不同的组合,所以我们要从不同数字的对应字母里分别找出一个字母进行组合,这个就和leetcode里combination的题目很相似,应该用recursion来解 解题思路如下: 比如说我们输入23: 那么按照我们的习惯, 我们从2中的abc选一个a 再...
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...
copy() else: num = 0 # combination for i in range(len(last)): for j in range(len(list_temp)): last_new[num] = last[i] + list_temp[j] num = num + 1 cur_y = cur_y + 1 last = last_new.copy() dyn_loop(list2,cur_y, last) return last Results:...
Letter Combinations of a Phone Number.go code算法 版权声明:原创勿转 https://blog.csdn.net/anakinsun/article/details/88867623 anakinsun 2019/04/12 4490 LeetCode 216. Combination Sum III(DFS) sum递归集合 题意:从1-9中选出k个数之和等于n,这个k个数不能有相同的,输出所有可能的k个数字的集合...
1publicclassSolution {2privatestaticfinalString[] KEYS = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};34publicList<String>letterCombinations(String digits) {5List<String> res =newLinkedList<String>();6combination("", digits, 0, res);7returnres;8}91...
Given a string containing digits from2-9inclusive, return all possible letter combinations that the number could represent. Return the answer inany order. A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters. ...
这道题让我们求电话号码的字母组合,即数字2到9中每个数字可以代表若干个字母,然后给一串数字,求出所有可能的组合,相类似的题目有Path Sum II 二叉树路径之和之二,Subsets II 子集合之二,Permutations 全排列,Permutations II 全排列之二,Combinations 组合项,Combination Sum 组合之和和Combination Sum II 组合之和...
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, ...
class LetterCombinationsPhoneNumber { func letterCombinations(_ digits: String) -> [String] { guard digits.count > 0 else { return [String]() } var combinations = [String](), combination = "" let numberToStr = ["", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv",...
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.