LeetCode 40. Combination Sum II

1. 题目描述
   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.
   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]
   ]

2. 解题思路
   这个题目就是背包问题简化版吧，二叉树，两种情况，放还是不放。
   一开始就是这样些的，后来发现一个小问题就是结果可能没有去重，如果背包有两个相同元素。
   后来我就很暴力的用set。。。。
   最后在转成vector返回。。。

3. 代码实现

#include<set>using namespace std;class Solution {public:    set<vector<int> > res;    int path[10000];    void getRes(vector<int>& nums,int target,int index,int sum,int pathlen){        if(sum==target){            vector<int> t;            for(int i=0;i<pathlen;++i)                t.push_back(path[i]);            res.insert(t);        }        else if(sum>target)            return ;        else {            if(index!=nums.size()){                //放进去                path[pathlen]=nums[index];                getRes(nums,target,index+1,sum+nums[index],pathlen+1);                //不放进去                //if(nums[index]!=nums[index+1])                getRes(nums,target,index+1,sum,pathlen);            }        }    }    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        getRes(candidates,target,0,0,0);        vector<vector<int> > vres;        set<vector<int>>::iterator it=res.begin();        for(;it!=res.end();++it)            vres.push_back(*it);        return vres;    }};