【leetcode】Combination Sum II (middle) ☆

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

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] 

思路：

把重复的数字看做一个整体，只能出现0-重复次数遍。 这个代码特别慢，不知道为什么 550ms

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target)  {
        vector<vector<int>> ans;
        if(num.empty())
            return ans;

        sort(num.begin(), num.end()); //从小到大排序
        recursion(ans, num, 0 , target);
        return ans;
    }

    void recursion( vector<vector<int> > &ans, vector<int> candidates, int k, int target)
    {
        static vector<int> partans;

        if(target == 0) //如果partans中数字的总和已经达到目标， 压入答案
        {
            ans.push_back(partans); 
            return;
        }
        if(k >= candidates.size() || target < 0)
            return;


        int num = candidates[k];
        int copy = 0;
        while(k < candidates.size() && candidates[k] == num)
        {
            k++;
            copy++;
        }

        recursion(ans, candidates, k, target);  //不压入当前数字

        for(int i = 1; i <= copy; i++)
        {
            partans.push_back(num); //压入当前数字
            recursion(ans, candidates, k , target - i * num); //后面只压入大于当前数字的数，避免重复
        }

        //恢复数据
        while(!partans.empty() && partans.back() == num)
        {
            partans.pop_back();
        }

    }
};

大神的13ms代码，感觉和我的差距不大，为什么速度快这么多。

class Solution {
    vector <int> path;
    vector < vector <int> > res;
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        sort(num.begin(), num.end());
        gen(0, target, num);
        return res;
    }
    void gen(int index, int sum, vector <int> &nums) {
        if (sum == 0) {
            res.push_back(path);
            return;
        }

        for (int i = index; i < nums.size(); i++) { //只压入序号大于等于i的数字 避免重复
            if (sum - nums[i] < 0) return;
            if (i && nums[i] == nums[i - 1] && index < i) continue; //每次递归相同的数字只压入一次
            path.push_back(nums[i]); //这里不需要不压入nums[i]的情况，因为循环到后面时自然就是未压入该数的情况了
            gen(i + 1, sum - nums[i], nums);
            path.pop_back();
        }
    }
};