Combination Sum II

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]

class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>>res;
sort(candidates.begin(),candidates.end());
if(candidates.empty()||target<candidates.front()) //数组为空或者是target小于最小的数组元素
return res;
vector<int>slo;
compute(0,target,candidates,slo,res);
return res;
}

void compute(int startIndex,int target,vector<int>&candidates,vector<int>&slo,vector<vector<int>>&res)
{
if(target<0)
return;
if(target==0)
{
res.push_back(slo);
return;
}
for(int i=startIndex;i<candidates.size();i++)
{

//消除在数组中重复出现的数字
if(i>startIndex&&candidates[i]==candidates[i-1]) continue;
if(target>=candidates[i])
{
slo.push_back(candidates[i]);
compute(i+1,target-candidates[i],candidates,slo,res);
slo.pop_back();
}
}
}
};