Leetcode 17 Letter Combinations of a Phone Number

Letter Combinations of a Phone Number

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"].

Solution1

• 求字符串之间的所有组合问题是典型的回溯问题。基本思路就是有几个数字便递归调用几轮，在每一轮调用的时候，去遍历所有的情况，如此递归到最深处。然后再回溯。代码如下：

import java.util.*;public class Solution {    public List<String> letterCombinations(String digits) {        char[][] table = {{'\0'},{'\0'},{'a','b','c'},{'d','e','f'},{'g','h','i'},                {'j','k','l'},{'m','n','o'},{'p','q','r','s'},                {'t','u','v'},{'w','x','y','z'}};        ArrayList<String> result = new ArrayList<String>();        help(digits,0,table,result,new StringBuffer());        return result;    }    public void help(String digits,int index,char[][] table,List<String> result,StringBuffer sb){        if(index>=digits.length()){            if(sb.length()!=0) result.add(sb.toString());            return;        }        int digit = digits.charAt(index) - '0';        for(int i=0;i<table[digit].length;i++){            StringBuffer nsb = new StringBuffer(sb);            if(digit!=0&&digit!=1) nsb.append(table[digit][i]);            help(digits,index+1,table,result,nsb);        }            }}

Solution2

• 下面用迭代的解法来求解这个问题。

import java.util.*;public class Solution {    public List<String> letterCombinations(String digits) {        String[] table = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};        ArrayList<String> result = new ArrayList<String>();        if(digits.length()==0) return result;        result.add("");        for(int i=0;i<digits.length();i++){            int digit = digits.charAt(i)-'0';            int size = result.size();//由于后边要更改result，所以必须得提前保存好当前result的长度            for(int k=size;k>0;k--){//每次都移除头部元素，移除k次                String s = result.get(0);                if(digit!=0&&digit!=1) result.remove(0);//注意只有在非0和非1的时候才将之前的移除                for(int j=0;j<table[digit].length();j++){                    String temp = new String(s);                    temp += table[digit].charAt(j);                    result.add(temp);                }            }        }        return result;         }}