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]
]
思路
本题是典型的回溯法，基本思路就是排序，回溯，但是要注意，数值可以被使用多次，所以回溯时要注意代码中i不变
backtrack(candidates,cur,i,target-candidates[i]);
代码
class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<int> cur;
sort(candidates.begin(),candidates.end());
backtrack(candidates,cur,0,target);
return res;
}
private:
void backtrack(vector<int> candidates,vector<int> cur,int start ,int target)
{
if(target==0)
{
res.push_back(cur);
}
else
{
for(int i=start;i<candidates.size()&&candidates[i]<=target;i++)
{
cur.push_back(candidates[i]);
backtrack(candidates,cur,i,target-candidates[i]);
cur.pop_back();
}
}
}
};
