[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];
        }
    }
};