0039_Combination Sum_c语言given a set of candidate numbers (candidates) (-优快云博客

本文链接：https://blog.youkuaiyun.com/wsy2846513/article/details/76473577

本文介绍了一种解决LeetCode上组合总和问题的方法，采用递归算法找到所有可能的数字组合，使得这些数字之和等于给定的目标值。通过实例演示了如何避免重复解并确保解集的唯一性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a set of candidate numbers (C) (without duplicates) 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]
]

JAVA

方法一

刚开始想使用hashmap来做，但是发现题中要求组合唯一，这样一来用hashmap出来的结果一定会有重复的，那么就要进行去重操作，而我又不会做。。
所以使用递归的方式，注意边界条件以及第一次调用时需要使用循环，以便可以从任意一个下标作为起始位置，否则会丢解。本来只是用这种暴力方式来试试，没想到效率在前1/4。莫非在LeetCode中递归的效率特别高？？？

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result = new LinkedList<List<Integer>>();
        LinkedList<Integer> tempResult = new LinkedList<Integer>();
        if(candidates.length != 0){
            Arrays.sort(candidates);
            for (int i = 0; i < candidates.length; i++) {
                getResult(candidates,target,result,i,0,tempResult);
            }
        }
        return result;
    }
    public void getResult(int[] candidates, int target,List<List<Integer>> result,
                          int currentIndex,int currentSum,LinkedList<Integer> tempResult){
        currentSum += candidates[currentIndex];
        tempResult.add(candidates[currentIndex]);
        if(currentSum == target){
            result.add((LinkedList)tempResult.clone());
        }
        while(currentIndex < candidates.length && currentSum + candidates[currentIndex] <= target){
            getResult(candidates,target,result,currentIndex,currentSum,tempResult);
            ++currentIndex;
        }
        tempResult.removeLast();
    }
}