【Leetcode】Letter Combinations of a Phone Number

问题

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.

代码

void myletterCombinations(const char *s , vector<string> &v , string currentString)
{
	if (!s) {
		return;
	}
	
	if (*s != '\0') {
		char testA[12];
		char a = ('a' + ((*s-'2') * 3));
		//sprintf(testA, "%c" , a + 4);
		/* get next */
		switch (*s) {
			case '9':
				myletterCombinations(s + 1, v, currentString + string("w"));
				myletterCombinations(s + 1, v, currentString + string("x"));
				myletterCombinations(s + 1, v, currentString + string("y"));
				myletterCombinations(s + 1, v, currentString + string("z"));
				break;
			case '8':
				myletterCombinations(s + 1, v, currentString + string("t"));
				myletterCombinations(s + 1, v, currentString + string("u"));
				myletterCombinations(s + 1, v, currentString + string("v"));
				break;
			case '7':
				myletterCombinations(s + 1, v, currentString + string("p"));
				myletterCombinations(s + 1, v, currentString + string("q"));
				myletterCombinations(s + 1, v, currentString + string("r"));
				myletterCombinations(s + 1, v, currentString + string("s"));
				break;
			case '6':
			case '5':
			case '4':
			case '3':
			case '2':
			{
				sprintf(testA, "%c" , a);
				myletterCombinations(s + 1, v, currentString + string(testA));
				sprintf(testA, "%c" , a+1);
				myletterCombinations(s + 1, v, currentString + string(testA));
				sprintf(testA, "%c" , a +2);
				myletterCombinations(s + 1, v, currentString + string(testA));
				break;
			}
			default:
				myletterCombinations(s + 1, v, currentString);
				break;
		}
	}
	else{
		v.push_back(currentString);
	}
	
}

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> ss;
		myletterCombinations(digits.c_str(), ss, string(""));
		return ss;
    }
};

分析

这里注意不仅仅是 9 有四个字符，7也是，我开始没注意。 2-6 是没问题的，使用通用的算法计算就可以了。 7,9 不行，8自然在中间也不能独自计算。

递归之就行了。跳过了 0 , 1, #,* 等无意义的输入，我没有试过假设不处理这几个字符是否会拒绝我的代码。

总结

递归各种组合。递归之前想好递归的停止条件是啥。