40. Combination Sum II

Problem:

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

Each number in candidates 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.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

思路：

和39题类似的思路，利用递归求解。注意要除去重复的结果。

Solution (C++):

public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> res;
        vector<int> combination;
        combinationSum2(candidates, target, res, combination, 0);
        
        return res;
    }
private:
    void combinationSum2(vector<int> &candidates, int target, vector<vector<int>> &res, vector<int> &combination, int begin) {
        if (target == 0) {
            res.push_back(combination);
            return;
        }
        else {
            for (int i = begin; i < candidates.size() && target >= candidates[i]; ++i) {
                if (i && candidates[i] == candidates[i-1] && i > begin)  continue;        //检查是否重复
                combination.push_back(candidates[i]);
                combinationSum2(candidates, target-candidates[i], res, combination, i+1);
                combination.pop_back();
            }
        }
    }

性能：

Runtime: 8 ms  Memory Usage: 8.9 MB