LeetCode 39. Combination Sum

最新推荐文章于 2025-12-10 14:25:37 发布

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本文详细解析了给定一组候选数，在没有重复的情况下找到所有可能的组合，使得这些组合的和等于目标值的问题。介绍了两种解决方案，一种是通过递归实现的直接回溯法，另一种是优化后的递归算法。

Given a set ofcandidate numbers (C) (without duplicates) and atarget number (T), find all unique combinations in C wherethe candidate numbers sums to T.

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

Note:

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

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

[

[7],

[2,2, 3]

]

直接回溯：

遍历数组每个元素：选择当前元素---》递归，不选择当前元素---》递归

再加上累计和（sum）与目标（target）比较判断

class Solution {
         List<List<Integer>>ans;
         inttarget;
   public List<List<Integer>> combinationSum(int[] candidates,int target) {
       ans = new ArrayList<List<Integer>>();
       this.target = target;
       Arrays.sort(candidates);
       travelCandidates(candidates,0,0,new ArrayList<Integer>());
       return ans;
    }
 
   public void travelCandidates(int[] candidates,int cur,intsum,List<Integer> list){
             if(cur>=candidates.length) return;
 
             travelCandidates(candidates,cur+1,sum,newArrayList<Integer>(list));
 
             sum += candidates[cur];
             list.add(candidates[cur]);
             if(sum == target){ans.add(list);return;}
             if(sum < target){travelCandidates(candidates,cur,sum,list);}
    }
}

提交后只超越了5%的提交者…看了下讨论模块，发现自己递归优化少（能用循环的就少用递归），优化代码：

public class Solution {
   public List<List<Integer>> combinationSum(int[] candidates,int target) {
             Arrays.sort(candidates);
       List<List<Integer>> result = newArrayList<List<Integer>>();
       getResult(result, new ArrayList<Integer>(), candidates, target,0);
        
       return result;
    }
   
   private void getResult(List<List<Integer>> result,List<Integer> cur, int candidates[], int target, int start){
             if(target > 0){
                       for(int i = start; i <candidates.length && target >= candidates[i]; i++){
                                cur.add(candidates[i]);
                                getResult(result,cur, candidates, target - candidates[i], i);
                                cur.remove(cur.size()- 1);
                       }//for
             }//if
             else if(target == 0 ){
                       result.add(newArrayList<Integer>(cur));
             }//else if
    }
}