40. Combination Sum II

Posted 2021-01-13 ruruozhenhao

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]
]

 

AC code:

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        set<vector<int>> s;
        vector<int> combination;
        vector<vector<int>> res;
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, s, combination, target, 0);
        set<vector<int>>::iterator it;
        for (it = s.begin(); it != s.end(); ++it)
            res.push_back(*it);
        return res;
    }
    
    void backtracking(vector<int>& candidates, set<vector<int>>& s, vector<int>& combination, int target, int begin) {
        if (!target) {
            s.insert(combination);
            return ;
        }
        for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {
            combination.push_back(candidates[i]);
            backtracking(candidates, s, combination, target-candidates[i], i+1);
            combination.pop_back();
        }
    }
};

Runtime: 8 ms, faster than 70.36% of C++ online submissions for Combination Sum II.