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39. Combination Sum(回溯)

ID amp number public ive turn mes without NPU

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]
]

 1 class
Solution { 2 private List<List<Integer>> res = new ArrayList<>(); 3 public List<List<Integer>> combinationSum(int[] candidates, int target) { 4 List<Integer> temp = new ArrayList<Integer>(); 5 help(temp,candidates,0,0,target); 6 return
res; 7 } 8 private void help(List<Integer> temp,int[] nums,int index,int cursum,int target){ 9 if(cursum>target) 10 return; 11 if(cursum==target) 12 res.add(new ArrayList<Integer>(temp)); 13 for(int i = index;i<nums.length;i++){ 14 temp.add(nums[i]); 15 help(temp,nums,i,cursum+nums[i],target); 16 temp.remove(temp.size()-1); 17 } 18 } 19 }

39. Combination Sum(回溯)