> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/graph-search/39.combination-sum.md).

# 39.Combination-Sum

## 39. Combination Sum

## 题目地址

<https://leetcode.com/problems/combination-sum/>

## 题目描述

```
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]
]
```

## 代码

### Approach 1: backtracking

```java
class Solution {
  public List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<List<Integer>> result = new ArrayList<>();
    if (candidates == null)    return result;

    List<Integer> combination = new ArrayList<>();
    Arrays.sort(candidates); // sort
    dfs(candidates, 0, target, combination, result);

    return result;
  }

  private void dfs(int[] candidates, int index, int target, List<Integer> combination, List<List<Integer>> result) {
    if (target == 0) {
      result.add(new ArrayList<Integer>(combination));
      return;
    }

    for (int i = index; i < candidates.length; i++) {
      if (candidates[i] > target) {
        break;
      }

      //  去重
      // if (i != 0 && candidates[i] == candidates[i - 1]) {
      //   continue;
      // } 

      combination.add(candidates[i]);
      dfs(candidates, i, target - candidates[i], combination, result);
      combination.remove(combination.size() - 1);
    }

  }
}
```

### Approach #2  Reuse

```java
public class Solution {
  public List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<List<Integer>> result = new ArrayList<>();
    if (candidates == null)    return result;

    dfs(candidates, 0, target, new ArrayList<Integer>(), result);
    return result;
  }

  private void dfs(int[] candidates, int index, int target, List<Integer> combination, List<List<Integer>> result) {
    if (target == 0) {
      result.add(new ArrayList<Integer>(combination));
      return;
    }
    if (index == candidates.length || target < 0) {
      return;
    }

    dfs(candidates, index + 1, target, combination, result);

    if (index > 0 && candidates[index] == candidates[index - 1]) {
      return;
    }

    combination.add(candidates[index]);
    dfs(candidates, index, target - candidates[index], combination, result);
    combination.remove(combination.size() - 1);
  }

}
```


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter, and the optional `goal` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/graph-search/39.combination-sum.md?ask=<question>&goal=<endgoal>
```

`ask` is the immediate question: it should be specific, self-contained, and written in natural language.
`goal` is optional and describes the broader end goal you are ultimately trying to accomplish on behalf of the user. GitBook uses it to tailor the answer towards what is most useful for that goal.

The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.