LeetCode 17. Letter Combinations of a Phone Number--输入数字，每个数字对应手机键盘的字符，输出组合的字符串

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"].

import java.util.LinkedList;
import java.util.List;

public class Main {

    public List letterCombinations(String digits) {
        LinkedList ans = new LinkedList();
        if (digits == null || digits.length() == 0) {
            return ans;
        }
        String[] intToStringArray = new String[]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        int x, n = digits.length();
        String s;
        ans.add("");
        for (int i = 0; i < n; i++) {
            while (ans.peek().length() == i) {//当ans首结点长度为0时，把第一个数代表的字符依次加入链表，
                // 当首结点长度为1时，需要进行下次循环，依次弹出首结点，首结点再与后面的数代表的每个字符累加，加入到链表中
                // 此时首结点的长度为2，重复操作..........最后输出
                s = ans.remove();//先获取链表的首结点赋值到s，再删除首结点，避免下次再循环到，下次需要用到新的首结点
                x = Character.getNumericValue(digits.charAt(i));//char->int
                for (char c : intToStringArray[x].toCharArray()) {//用弹出的首结点去累加后面的字符
                    ans.add(s + c);
                }
            }
        }
        return ans;
    }//letterCombinations


    public static void main(String[] args) {
        System.out.println(new Main().letterCombinations("02"));//[0a, 0b, 0c]
        System.out.println(new Main().letterCombinations("234"));//[adg, adh, adi, aeg, aeh, aei, afg, afh, afi, bdg, bdh, bdi,
        // beg, beh, bei, bfg, bfh, bfi, cdg, cdh, cdi, ceg, ceh, cei, cfg, cfh, cfi]
//        System.out.println(new Main().letterCombinations(""));

    }
}

//peek是返回首结点的值，不删除首结点
//remove是返回首结点的值，删除首结点