LeetCode进阶之路（Letter Combinations of a Phone Number)

本文链接：https://blog.youkuaiyun.com/yulingkai88/article/details/51935892

本文介绍了一种通过递归深度优先搜索解决电话号码对应字母组合问题的方法。给出数字字符串，返回所有可能的字母组合。文章提供了详细的代码实现，并附上了完整的递归函数。

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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.

这道题目实在没思路，感觉暴力解法也需要循环若干次，参考网友的做法，贴上他博客的地址：

http://blog.youkuaiyun.com/china_wanglong/article/details/38495355

虽然看懂，但是感觉自己肯定想不到这个思路，这也许就是所谓思维的局限性，还有好多的东西要学。。直接贴上他的解法，有空在慢慢研究

public List<String> letterCombinations(String digits) {  
        List<String> result = new ArrayList<String>();  
        String[] map = new String[10];  
        map[0] = "";  
        map[1] = "";  
        map[2] = "abc";  
        map[3] = "def";  
        map[4] = "ghi";  
        map[5] = "jkl";  
        map[6] = "mno";  
        map[7] = "pqrs";  
        map[8] = "tuv";  
        map[9] = "wxyz";  
        char[]  middleTemp = new char[digits.length()];  
        dfsGetStr(digits, 0, middleTemp, map, result);   
        return result;  
    }  
      
    private void dfsGetStr(String digits,int index,   
            char[] middleStr, String[] map, List<String> result) {  
        if(index == digits.length()) {  
            result.add(new String(middleStr));  
            return ;  
        }  
        char strChar = digits.charAt(index);  
        for(int i=0; i<map[strChar-'0'].length(); i++) {  
            middleStr[index] = map[strChar-'0'].charAt(i);  
            dfsGetStr(digits, index+1, middleStr, map, result);  
        }  
    }