LeetCode 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. Return the answer in any order.

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 1:

复制Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]

Example 2:

Input: digits = ""
Output: []

Example 3:

Input: digits = "2"
Output: ["a","b","c"]

Constraints:

0 <= digits.length <= 4
digits[i] is a digit in the range ['2', '9'].

实现思路：

题目要求给出一组数字按键，然后不同数字按键有一组不同的字符，要求输出所有按键能够组成的不同字符组合。
这种要求不同字符组成不同的组合的题目首先应该想到的就是回溯算法，利用全排列算法作为基础模型，当然本题有点类似于N-皇后问题，不妨先写出N-皇后那种方式，就是2~9所有不同按键中的字符能够组成的不同字符情况。

代码如下：

class Solution {//创建长度为9的随机串 

		vector<char> sq;
		vector<string> str= {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"},ans;

		void dfs(int x) {
			if(x>9) {
				string temp;
				for(int i=0; i<sq.size(); i++)
					temp+=sq[i];
				ans.push_back(temp);
				return;//递归边界
			}
			string temp;
			for(int i=0; i<str[x].size(); i++) {
				sq.push_back(str[x][i]);
				dfs(x+1);
				sq.pop_back();
			}
		}

	public:
		vector<string> letterCombinations(string digits) {
			dfs(2);
			return ans;
		}
};

这时候便开始思考，本题要求的是，如果给出“234”，那么这三个按键组成的字符长度就应该是3，所以我们必须要用题目输入string的长度作为递归边界来解答。

AC代码：

class Solution {
		vector<char> sq;
		vector<string> str= {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"},ans;
		void dfs(int x,string &s,int size) {
			if(x==size) {//当得到size个不同案件上的字符时候为递归边界
				string temp;
				for(int i=0; i<sq.size(); i++)
					temp+=sq[i];
				if(temp.size()>0)//避免空字符串存入
					ans.push_back(temp);
				return;
			}
			string temp;
			int k=s[x]-'0';//针对数组中每个数字找到对应按键对应的字母
			for(int i=0; i<str[k].size(); i++) {
				sq.push_back(str[k][i]);
				dfs(x+1,s,size);//x的作用是控制已经存入几个char字符
				sq.pop_back();
			}
		}

	public:
		vector<string> letterCombinations(string digits) {
			dfs(0,digits,digits.size());
			return ans;
		}
};