首页 > 代码库 > Combination Sum

Combination Sum

2024-07-28 09:16:27   222人阅读

Given a set of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums to T.

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

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1, a2, … ,ak) must be in non-descending order. (ie, a1 ≤a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

答案

public class Solution {
    int[] candidates;
    public List<List<Integer>> combinationSum(int index ,int target)
    {
        List<List<Integer>> result=new LinkedList<List<Integer>>();
        if(index>=candidates.length||candidates[index]>target)
        {
            return result;
        }
        List<Integer> p=new LinkedList<Integer>();
        while(target>=0)
        {
            if(target==0)
            {
                result.add(p);
                break;
            }
            List<List<Integer>> next=combinationSum(index+1,target);
            if(next.size()>0)
            {
                for(List<Integer> list:next)
                {
                    List<Integer> pList=new LinkedList<Integer>();
                    pList.addAll(p);
                    pList.addAll(list);
                    result.add(pList);
                }
            }
            p.add(candidates[index]);
            target-=candidates[index];
        }
        return result;
        
    }
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result=new LinkedList<List<Integer>>();
        if(candidates==null||target<=0)
        {
            return result;
        }
        Arrays.sort(candidates);
        this.candidates=candidates;
        return combinationSum(0,target);
    }
}


ArrayBacktracking

Combination Sum


联系
我们