0
点赞
收藏
分享

微信扫一扫

39. Combination Sum

Greatiga 2022-08-03 阅读 38


Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

public class Solution {  
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> ans = new ArrayList<>();
helper(0, target, 0, candidates, new ArrayList<>(), ans);
return ans;
}

private void helper(int index, int target, int sum, int[] candidates,
List<Integer> list, List<List<Integer>> ans)
{
if(index > candidates.length-1 || sum > target)
return;
if(sum == target && list.size() != 0)
ans.add(new ArrayList<>(list));
for(int i = index; i < candidates.length; i++) {
list.add(candidates[i]);
helper(i, target, sum+candidates[i], candidates, list, ans);
list.remove(list.size()-1);
}
}
}


举报

相关推荐

0 条评论