[leetcode][17] Letter Combinations of a Phone Number

17. Letter Combinations of a Phone Number

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.

解析

模拟手机按键，输入一串按下的数字，返回所有可能的字符串。

参考答案

自己写的：

class Solution {
    private static List<String> lists = new ArrayList<>();
    
    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) {
            return new ArrayList<>();
        }
        lists.add("abc");
        lists.add("def");
        lists.add("ghi");
        lists.add("jkl");
        lists.add("mno");
        lists.add("pqrs");
        lists.add("tuv");
        lists.add("wxyz");
        List<Integer> indexs = new ArrayList<>();
        for (int i = 0; i< digits.length(); i++) {
            indexs.add(Integer.valueOf(digits.charAt(i) + ""));
        }
        List<StringBuilder> sb = new ArrayList<>();
        sb.add(new StringBuilder());
        List<String> res = createResult(0, indexs, sb);
        return res;
    }
    
    public List<String> createResult(int i, List<Integer> indexs, List<StringBuilder> builders) {
        if (i == indexs.size()) {
            List<String> res = new ArrayList<>();
            for (StringBuilder sb : builders) {
                res.add(sb.toString());
            }
            return res;
        }
        List<StringBuilder> next = new ArrayList<>();
        for (StringBuilder sb : builders) {
            String s = this.lists.get(indexs.get(i)-2);
            for (int j = 0; j < s.length(); j++) {
                StringBuilder newSb = new StringBuilder(sb);
                newSb.append(s.charAt(j));
                next.add(newSb);
            }
        }
        i++;
        return createResult(i, indexs, next);
    } 
}

写的很复杂，用递归的方法，递归的层数等于按键的个数，每层的结果在上层的结果之上加上新的字符，到了最后一层返回。

别人写的（很简洁）：

public List<String> letterCombinations(String digits) {
        LinkedList<String> ans = new LinkedList<String>();
        if(digits.isEmpty()) return ans;
        String[] mapping = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        ans.add("");
        while(ans.peek().length()!=digits.length()){
            String remove = ans.remove();
            String map = mapping[digits.charAt(remove.length())-'0'];
            for(char c: map.toCharArray()){
                ans.addLast(remove+c);
            }
        }
        return ans;
    }
    

这里很巧妙的用到了队列，把上一个按键得到字符串，出队列加上新的字符，再入队列，还有一个巧妙的点在于：

String map = mapping[digits.charAt(remove.length())-'0']

用Ascii码的差值得到index，不用做类型转换。