LeetCode Combination Sum II

题目

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]

和上一题类似，只是元素不能重复取。

代码：

class Solution {
vector<vector<int>> ans;
vector<int> temp;
public:
void combs2(vector<int> &cand,int target,int sum,int begin)	//候选集合，目标，当前序列和，当前可以取的最小编号
{
if(sum==target)	//获得结果
{
ans.push_back(temp);
return;
}
for(int i=begin;i<cand.size();i++)	//查找
{
if(sum+cand[i]<=target)
{
temp.push_back(cand[i]);
combs2(cand,target,sum+cand[i],i+1);
temp.pop_back();
while(i+1<cand.size()&&cand[i]==cand[i+1])	//使要更改的位置可以取得不同的数，从而保证不出现重复解
i++;
}
else
break;
}
}
vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
ans.clear();
temp.clear();
if(candidates.empty())
return ans;
sort(candidates.begin(),candidates.end());
combs2(candidates,target,0,0);
return ans;
}
};