[leetcode]Combination Sum II

问题描述：

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums toT.

Each number in C may only be usedonce 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 target8,
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，只是加了一个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;    }    };