题目
Given a set of candidate numbers (C) 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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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
用递归的思想实现~
AC代码
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
if (candidates.empty() || target < 0)
return vector<vector<int> >();
ret.clear();
//将给定序列排序
sort(candidates.begin(), candidates.end());
vector<int> tmp;
combination(candidates, 0, tmp, target);
return ret;
}
//递归实现
void combination(vector<int> &candidates, int idx, vector<int> &tmp, int target)
{
if (target == 0)
{
ret.push_back(tmp);
return;
}
else{
int len = candidates.size();
for (int i = idx; i < len; i++)
{
if (target >= candidates[i])
{
tmp.push_back(candidates[i]);
combination(candidates, i, tmp, target - candidates[i]);
tmp.pop_back();
}//if
}//for
}//else
}
private:
vector<vector<int> > ret;
};