leetcode-letter combination of a phone number

来源:互联网 发布:制作微课都用哪些软件 编辑:程序博客网 时间:2024/06/01 07:53

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.

public class Solution {    //store the mapping    String[] map={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};    List<String> result=new ArrayList<String>();    public List<String> letterCombinations(String digits) {        if(digits.length()==0){            result.add(new String());            return result;        };      findCombination(digits,0,new StringBuffer());        return result;    }    public void findCombination(String digits,int pos, StringBuffer s){        int num=digits.charAt(pos)-'0';        int len=map[num].length(); // length of the string the digit refers to                  for(int i=0;i<len;i++){             s.append(map[num].charAt(i));             if(pos==digits.length()-1)result.add(s.toString());             else if(pos<digits.length()-1){findCombination(digits,pos+1,s);}             /*undo effect*/             s.deleteCharAt(pos);  //how to use         }    }}
想法非常简单:用string数组储存电话键盘上面的数字,注意对应的数组下标要和数字对应。

backtracking传递参数,第几个数字,string。

注意的问题:stringbuffer有toString()方法,builder没有。然后delete方法不要忘记。

2在没有输入的情况下传递一个包含空的字符串result。

3charAt ()返回来的是一个char 对应的二进制,不要忘了减去'0'转换成想要的数字。


0 0
原创粉丝点击