LeetCode 39. Combination Sum

最新推荐文章于 2025-05-25 18:32:47 发布

nianhua120

最新推荐文章于 2025-05-25 18:32:47 发布

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本文介绍了一种基于回溯遍历的递归算法，用于从给定候选数集中找到所有可能的组合，使得这些组合的元素之和等于目标值。算法允许候选数集中的相同数字被无限次使用。

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Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C 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.
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]
]

本题考查的是回溯遍历采用递归很容易求解。（注意避免重复）

public class Solution {
    private void helper(int[] candidates, int target, List<List<Integer>> list ,int curSum, List<Integer> curList, int s){
        if(curSum > target){
            return;
        }        
        for(int i=s; i<candidates.length; i++){
            curList.add(candidates[i]);
            if(curSum+candidates[i] == target){
                List subtle = new LinkedList<Integer>(curList);
                list.add(subtle);
            }else if(curSum+candidates[i] < target){
                helper(candidates, target, list, curSum+candidates[i], curList, i);
            }else{
             
            }
            curList.remove(curList.size()-1);
        }
    }
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> list = new LinkedList<>();
        List<Integer> curList = new LinkedList<>();
        helper(candidates, target, list, 0, curList, 0);
         return list;
    }
}