# [39. Combination Sum](https://leetcode.com/problems/combination-sum/)


## 题目

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]
    ]


## 题目大意

给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。


## 解题思路

- 题目要求出总和为 sum 的所有组合，组合需要去重。
- 这一题和第 47 题类似，只不过元素可以反复使用。

## 代码

```go

package leetcode

import "sort"

func combinationSum(candidates []int, target int) [][]int {
	if len(candidates) == 0 {
		return [][]int{}
	}
	c, res := []int{}, [][]int{}
	sort.Ints(candidates)
	findcombinationSum(candidates, target, 0, c, &res)
	return res
}

func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
	if target <= 0 {
		if target == 0 {
			b := make([]int, len(c))
			copy(b, c)
			*res = append(*res, b)
		}
		return
	}
	for i := index; i < len(nums); i++ {
		if nums[i] > target { // 这里可以剪枝优化
			break
		}
		c = append(c, nums[i])
		findcombinationSum(nums, target-nums[i], i, c, res) // 注意这里迭代的时候 index 依旧不变，因为一个元素可以取多次
		c = c[:len(c)-1]
	}
}

```