回溯法（leetcode-Combination Sum）

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7, 
A solution set is: 

[
  [7],
  [2, 2, 3]
]

本题宜采用回溯法，代码如下：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

void backTracking(vector<vector<int>> &res, vector<int> arr ,vector<int>candidate ,int start,int target )
{
	if (target == 0)//满足条件，输出结果
	{
		res.push_back(arr);
		return;
	}
	for (int i = start; i < candidate.size(); i++)//枚举所有可能的路径
	{
		if (candidate[i] <= target)//满足界限函数和约束条件
		{
			arr.push_back(candidate[i]);
			backTracking(res, arr, candidate, i, target - candidate[i]);
			arr.pop_back();//回溯清理工作
		}
	}
}

int main()
{
	vector<int> candidate = { 2, 3, 6, 7 };
	int target = 7;
	vector<vector<int>> res;
	vector<int> arr;
	sort(candidate.begin(), candidate.end());
	backTracking(res, arr, candidate, 0, target);
	for (int i = 0; i < res.size(); i++)
	{
		for (int j = 0; j < res[i].size(); j++)
		{
			cout << res[i][j] << " ";
		}
		cout << endl;
	}
	return 0;
}