39-Combination Sum

题目：

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]
]

分析：

和N-Queens && N-Queens II 的方法很像，采用递归回溯的方法。
之前接触的深度优先遍历一般是应用到二叉树的遍历，或者图的遍历。虽然已经做了两次这种回溯方法的问题，还是不是很适应这种思路。
递归的代码很简单，但是要自己想明白，写出来还是有很大的差距的。

实现：

class Solution {

private :
vector<vector<int>> res;

public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {

vector<int> tmp;
sort(candidates.begin(), candidates.end());

DFS(tmp, candidates,target, 0);

return res;

}

int sum(vector<int> tmp)
{
int result = 0;
for (int i = 0; i < tmp.size(); i++)
{
result += tmp[i];
}
return result;

}
//递归
void DFS(vector<int>& tm
4000
p, vector<int>& candidates, int target, int len)
{
if (len == candidates.size())
return;
if (target == sum(tmp))
{
res.push_back(tmp);
return;

}
else if (target < sum(tmp))
return;
else
{
//核心代码
for (int i = len; i < candidates.size(); i++)
{
tmp.push_back(candidates[i]);
//i 表示该数字可被重复使用
DFS(tmp, candidates, target, i);
tmp.pop_back();

}
}

}
};