39. Combination Sum (Java)

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]
]

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

    public void backTrack(int[] candidates, int target, int start, List<Integer> ans, int sum){
        if(start >= candidates.length || sum + candidates[start] > target)
            return; //not found
        else if(sum + candidates[start] == target ){ //found an answer
            List<Integer> new_ans = new ArrayList<Integer>(ans); //不能用List<Integer> new_ans = ans;这个只是创建了原List的一个引用
            new_ans.add(candidates[start]);
            result.add(new_ans);
        }
        else{
            // not choose current candidate
            backTrack(candidates,target,start+1,ans,sum);

            //choose current candidate
            ans.add(candidates[start]);
            backTrack(candidates,target,start,ans,sum+candidates[start]);
            ans.remove(ans.size()-1); //List是按引用传递，为了不影响递归，需要复原
        }
    }

    private List<List<Integer>> result = new ArrayList<List<Integer>>();
}

注意List按引用（地址）传递，需要新new一个，或是复原，以不影响其他部分

原文地址：https://www.cnblogs.com/qionglouyuyu/p/10812822.html