[leetcode]Combination Sum II

问题描写叙述：

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC 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]

基本思路：

该题与Combination Sum 类似，仅仅是加了一个base数组用于记录数字是否被使用过。

代码：

class Solution {   //C++
public:
vector<int> record;
vector<vector<int> >result;
set<vector<int> > myset;

void addSolution(){
vector<int> tmp(record.begin(),record.end());
sort(tmp.begin(),tmp.end());
if(myset.find(tmp) == myset.end()){
result.push_back(tmp);
myset.insert(tmp);
}
}
void subCombinationSum(vector<int> &cadidates, int target,int bpos,vector<int> &base){
if(target ==0 ){
addSolution();
}

int size = cadidates.size();
for(int i = bpos; i < size&&cadidates[i] <= target; i++){
if(base[i] == 1)
continue;

record.push_back(cadidates[i]);
base[i] = 1;
subCombinationSum(cadidates,target-cadidates[i],bpos,base);
record.pop_back();
base[i] = 0;
}
}

vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {
sort(candidates.begin(),candidates.end());
vector<int> base(candidates.size(),0);
subCombinationSum(candidates,target,0,base);
return result;
}

};