leetcode第17题——**Letter Combinations of a Phone Number

题目

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.

思路

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).

代码

Python

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

Java

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;    }}