Given a digit string, 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.
Input:Digit string "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
需要三层循环:最外层循环,遍历数字串,一个数字对应一个相应的字符串,如2对应“abc", 3对应"def";中间层循环,遍历数字对应的字符串,每个字符新加进res的字符串里;最里层循环,遍历已保存的res链表的所有字符串,每个字符串都加上一个字符。虽然看上去有三层循环嵌套,但中间层循环最多也就遍历四次(数字对应的字符串长度最大为4),所以算法的整体时间复杂度大概为O(n^2).
class Solution(object): def letterCombinations(self, digits): """ :type digits: str :rtype: List[str] """ if len(digits) == 0: return [] #按照键盘分布,下标0-9分别对应字符串 digitLis = ["0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"] res = [""] for num in digits: tempLis = [] for ch in digitLis[int(num)]: for str in res: tempLis.append(str + ch) res = tempLis return res
public class Solution { public List<String> letterCombinations(String digits) { List<String> res = new ArrayList<String>(); int len = digits.length(); if (len == 0) return res; //按照键盘分布,初始化一个字符串数组,下标0-9分别对应指定的字符串 String[] digitArr = {"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; res.add(""); int i,j,k; for (i = 0;i < len;i++){ List<String> tempLis = new ArrayList<String>(); String iStr = digitArr[digits.charAt(i) - '0'];//找出数字对应的字符串 for (j = 0;j < iStr.length();j++) for (k = 0;k < res.size();k++) tempLis.add(res.get(k) + iStr.charAt(j)); res = tempLis; } return res; } }