39. Combination Sum(medium, backtrack 的经典应用, 重要)

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 the 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]
]

本题是 backtrack(回溯法) 的经典应用.通过本题，仔细观察 backtrack 基本框架，并感受其应用.

另外, subset 的题目也用 backtrack 方法.

人家想法，自个代码:

• cpp 副产品： v.pop_back(); 弹出队尾元素.
• 要注意的是: 下面代码中 backtrack 方法中的 res, temp, A, 尤其是 temp, 他们都分别指向对应的一个对象，无论 backtrack 方法被嵌套地调用多少次！

// backtrace method
vector<vector<int>> combinationSum(vector<int>& A, int ta) {
	sort(A.begin(), A.end());
	vector < vector<int> > res;
	vector<int> temp;
	backtrack(res, temp, A, ta, 0);
	return res;
}

void backtrack(vector<vector<int> >& res, vector<int>& temp, vector<int>& A,
		int remain, int start) {
	if (remain < 0)
		return;
	else if (remain == 0) {
		res.push_back(temp);
		return;
	} else {
		for (int i = start; i < A.size(); i++) {
			temp.push_back(A[i]);

			// not i + 1, because A[i] might be reused.
			backtrack(res, temp, A, remain - A[i], i);

			temp.pop_back();
		}
		return;
	}
}