200112题（很不错的一道回溯法题）

这道题要经常拿出来重新看一看，动手编一编！

和leetcode第39题组合总和的思路类似。

思路：回溯法

#include<iostream>
#include<map>
#include<vector>
#include<iterator>
#include<string>
using namespace std;

map<char, string> mp = { { '2',"abc" },{ '3',"def" },{ '4',"ghi" },{ '5',"jkl" },{ '6',"mno" },{ '7',"pqrs" },{ '8',"tuv" },{ '9',"wxyz" } };
class Solution {
public:
	vector<string> letterCombinations(string numberstr) {
		if (numberstr.size() == 0)
			return res;
		DFS(numberstr);
		return res;
	}
	void DFS(string numberstr)
	{
		if (numberstr.size() == 0)//递归结束的条件
		{
			res.push_back(cur);
			return;
		}
		string letterstr = mp[numberstr[0]];

		for (int i = 0; i < letterstr.size(); i++) {
			cur.push_back(letterstr[i]);
			DFS(numberstr.substr(1));//substr函数返回的是对象，不是引用
			cur.pop_back();//核心：上一句总是会出来的
		}

	}
private:
	vector<string> res;
	string cur;
};

int main() {
	Solution s;
	vector<string>result = s.letterCombinations("23");
	copy(result.begin(), result.end(), ostream_iterator<string>(cout, " "));
	system("pause");
	return 0;
}