[LeetCode]17. Letter Combinations of a Phone Number

Problem Description

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.
[https://leetcode.com/problems/letter-combinations-of-a-phone-number/]

思路

递归循环构造字符串即可…
第二次一遍AC 鼓励自己一下 看会死神去咯~

Code

package Q17;import java.util.ArrayList;import java.util.List;public class Solution {    public static List<String> getLetter(List<String> tmp, String digits) {        List<String> ans = new ArrayList<String>();        if (tmp.size() > 0) {            for (int i = 0; i < digits.length(); i++) {                for (int j = 0; j < tmp.size(); j++) {                    ans.add(tmp.get(j) + digits.charAt(i));                }            }        }        else{            for (int i = 0; i < digits.length(); i++) {                    ans.add(digits.charAt(i)+"");            }        }        return ans;    }    public static List<String> letterCombinations(String digits) {        List<String> ans = new ArrayList<String>();        List<String> digits2letters = new ArrayList<String>();        digits2letters.add(" ");        digits2letters.add("");        digits2letters.add("abc");        digits2letters.add("def");        digits2letters.add("ghi");        digits2letters.add("jkl");        digits2letters.add("mno");        digits2letters.add("pqrs");        digits2letters.add("tuv");        digits2letters.add("wxyz");        if (digits.length() < 1)            return ans;        for (int i = 0; i < digits.length(); i++) {            if(digits.charAt(i) - 48==1||digits.charAt(i) - 48==0) return new ArrayList<String>();            ans = getLetter(ans, digits2letters.get(digits.charAt(i) - 48));        }        return ans;    }//  public static void main(String[] args) {//      String a = "123";//      System.out.println(letterCombinations(a).toString());//  }}