[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, … ,a_k) must be in non-descending order. (ie, a₁ ≤a₂ ≤ … ≤ a_k).
• 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;
    }
    
};