Letter Combinations of a Phone Number：深度优先和广度优先两种解法

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"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

解法一：深度优先解法

即backtracking方法，每次向下搜索直到触底。这种方法采用LIFO（后进后出），通过递归实现。

public class Solution {
    public List<String> letterCombinations(String digits) {
        LinkedList<String> rst = new LinkedList<String>();
        if (digits == null || digits.length() == 0) {
            return rst;
        }
        String[] mapping = new String[]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        String temp = "";
        helper(rst, temp, digits, mapping, 0);
        return rst;
    }
    public void helper(List<String> rst, String temp, String digits, String[] mapping, int index) {
        if (index == digits.length()) {
            rst.add(new String(temp));
            return;
        }
        String s = mapping[Character.getNumericValue(digits.charAt(index))];
        for (int i = 0; i < s.length(); i++) {
            temp += s.charAt(i);
            helper(rst, temp, digits, mapping, index + 1);
            temp = temp.substring(0, temp.length() - 1);
        }
    }
}

解法二：广度优先解法

这种方法采用FIFO（先进先出），通过非递归方式实现。具体使用了LinkedList数据结构的add方法在list尾部插入，remove方法在list头部删除来实现FIFO。此外，还巧妙地运用了peek方法，每次新加元素时更新原有的每个结果。

public class Solution {
    public List<String> letterCombinations(String digits) {
        LinkedList<String> rst = new LinkedList<String>();
        if (digits == null || digits.length() == 0) {
            return rst;
        }
        String[] mapping = new String[]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        rst.add("");
        for (int i = 0; i < digits.length(); i++) {
            int m = Character.getNumericValue(digits.charAt(i));
            while(rst.peek().length() == i) {
                String s = rst.remove();
                for (char c: mapping[m].toCharArray()) {
                    rst.add(s + c);
                }
            }
        }
        return rst;
    }
}

值得思考的是，从直观上看第二种方法非递归应该运行效率更高，但实际两种方法的运行时间是一样的，都打败了46.02%的java submissions。