leetcode 39) Combination Sum

LV. Medium 🧐

https://leetcode.com/problems/combination-sum/

Combination Sum - LeetCode

 

문제

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

 

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

 

Constraints:

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• All elements of candidates are distinct.
• 1 <= target <= 500

 

문제 해결법

조합들을 중복있게 만드는 문제이다.

백트래킹을 사용해서 풀었다.

 

해결 코드

class Solution {
private:
	void sum(vector<int> &candidates, vector<vector<int>> &ans, vector<int> &list, int target, int i) {
		if (target == 0) {
			ans.push_back(list);
			return ;
		}
		while (i < candidates.size() && target - candidates[i] >= 0) {
			list.push_back(candidates[i]);
			sum(candidates, ans, list, target - candidates[i], i);
			++i;
			list.pop_back();
		}
	}
public:
	vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
		vector<vector<int>> ans;
		vector<int> list;

		sort(candidates.begin(), candidates.end());
		candidates.erase(unique(candidates.begin(), candidates.end()), candidates.end());
		sum(candidates, ans, list, target, 0);
		return ans;
	}
};