[LeetCode] Combination Sum II

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.
• 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]

以下是我 AC 的代码：

/**

 * author: Zhou J

 */

class Solution 

{

public:

    vector<vector<int> > combinationSum2(vector<int> &num, int target) 

    {

        sort(num.begin(), num.end());

        vector<vector<int>> ret;

        vector<int> path;

        combinationSum2Rec(num, target, ret, path, 0);

        return ret;

    }

    

private:

    void combinationSum2Rec(const vector<int> &num,

                            const int target,

                            vector<vector<int>> &ret,

                            vector<int> &path,

                            size_t pos)

    {

        if (target == 0)

        {

            ret.push_back(path);

            return;

        }

        

        // Mark the previous one

        int prev = -1;

        

        for (size_t ix = pos; ix != num.size(); ++ix)

        {

            if (num[ix] > target)

            {

                break;

            }

            

            if (prev == num[ix])

            {

                continue;

            }

            

            path.push_back(num[ix]);

            combinationSum2Rec(num, target - num[ix], ret, path, ix + 1);

            // Backtracing. Note that we should sort the vector before we use the skill of backtracing.

            path.erase(path.end() - 1);

            // The one which we have pushed just now. 

            prev = num[ix];

        }

    }

};