[LeetCode]17. Letter Combinations of a Phone Number电话号码的字母组合

Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.

Example:

Input: "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.

这题要求列出所有可能古德字符组合，我们先把1-9的字符列出来，然后从digits中一次取出数字代表的字符，然后做笛卡尔积的处理

这里的难点在于2个数的时候好处理，2层的循环就能解决，但是3个数，4个数怎么做呢？
4个数时，我们需要先求出2个数的笛卡尔积，然后再去处理第3个，第4个

所以我们在2层循环上，加上一层控制，在外部选定完i后，我们要在原来的字符串基础上加上第i个数字代表的字符，我们每次把list(0)的元素remove出来，然后加上

新的字符再添加到list的末尾，这样我们就可以根据list(0)的长度判断这一轮字符有没有全部加上，就像[b,c,ad,ae,af]这样，此时i=1,list.get(0).length()=1,这就说明还有这轮还没加完

直到[ad, ae, af, bd, be, bf, cd, ce, cf],此时i=1但list.get(0).length()=2，说明这轮结束，回到外层

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> list = new ArrayList<String>();
        if(digits.isEmpty()) return list;
        String[] str = new String[] {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        list.add("");
        String temp = null;
        for(int i = 0; i < digits.length(); i++) {
            int k = Integer.parseInt(digits.charAt(i)+"");
            while(list.get(0).length() == i)
            {
                 temp = list.remove(0);
                 for(int j = 0; j < str[k].length(); j++){
                     list.add(temp + str[k].charAt(j));
                  }
            }
        }
        return list;
    }
}