Java for LeetCode 017 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.

解题思路：

使用DFS算法，JAVA实现如下：

	    static String[] alpha = new String[] {
	    		" ","1", "abc", "def","ghi", "jkl", "mno","pqrs", "tuv", "wxyz"
	    };
	    static StringBuilder sb = new StringBuilder();
	    static  void dfs(List<String> list, String digits, int cur) {
	        if (cur >= digits.length())
	            list.add(sb.toString());
	        else {
	            for (int i = 0; i < alpha[digits.charAt(cur) - ‘0‘].length(); i++) {
	                sb.append(alpha[digits.charAt(cur) - ‘0‘].charAt(i));
	                dfs(list, digits, cur + 1);
	                sb.deleteCharAt(sb.length() - 1);
	            }
	        }
	    }
	   static public List<String> letterCombinations(String digits) {
	        List<String> list = new ArrayList<String>();
	        if (digits.length()==0)
	        　　return list;
	        dfs(list, digits, 0);
	        return list;
	    }

思路二：

凡是用到递归的地方都能用循环解决，因此可以用循环算法，JAVA实现如下：

static public List<String> letterCombinations(String digits) {
	        List<String> list = new ArrayList<String>();
	        String[] alpha = new String[] {
		    		" ","1", "abc", "def","ghi", "jkl", "mno","pqrs", "tuv", "wxyz"
		    };
	        if (digits.length()==0)
	        	return list;
	        int[] number = new int[digits.length()];//存储每次遍历字符位置
	        int index = 0;
	        while(index>=0) {
	            StringBuilder sb = new StringBuilder();
	            for(int i=0; i<digits.length(); i++)
	                sb.append(alpha[digits.charAt(i)-‘0‘].charAt(number[i]));
	            list.add(sb.toString());
	            // 每回合需要重置index到末尾
	            index = digits.length()-1;
	            while(index>=0) {
	                if( number[index] < (alpha[digits.charAt(index)-‘0‘].length()-1) ) {
	                    number[index]++;
	                    break;
	                } else {
	                    number[index] = 0;
	                    index--;
	                }
	            }
	        }
	        return list;
	    }