Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums to T.
Each number in C may only be used once in the combination.
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 10,1,2,7,6,1,5 and
target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
#include<vector>
#include<algorithm>
#include<numeric>
#include<set>
using namespace std;
void FindcombinationSum(vector<vector<int> >&ResultVector, vector<int> &candidates, vector<int> &Oneresult, int target, int num = 0)
{
if (accumulate(Oneresult.begin(), Oneresult.end(), 0) >= target)
return;
for (int i = num; i != candidates.size();++i)
{
if (candidates[i] == candidates[i – 1] && i>num)//保证不前1个重复数字必须被使用
continue;
Oneresult.push_back(candidates[i]);
int sum = accumulate(Oneresult.begin(), Oneresult.end(), 0);
if (sum == target)
ResultVector.push_back(Oneresult);
FindcombinationSum(ResultVector, candidates, Oneresult, target, i + 1);
Oneresult.pop_back();
}
}
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int> > ResultVector;
vector<int> Oneresult;
sort(candidates.begin(), candidates.end());
FindcombinationSum(ResultVector, candidates, Oneresult, target);
return ResultVector;
}
波比源码 » Combination Sum II
