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88 lines
1.9 KiB
Markdown
Executable File
88 lines
1.9 KiB
Markdown
Executable File
# [39. Combination Sum](https://leetcode.com/problems/combination-sum/)
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## 题目
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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`.
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The **same** repeated number may be chosen from `candidates` unlimited number of times.
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**Note**:
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- All numbers (including `target`) will be positive integers.
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- The solution set must not contain duplicate combinations.
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**Example 1**:
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Input: candidates = [2,3,6,7], target = 7,
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A solution set is:
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[
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[7],
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[2,2,3]
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]
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**Example 2**:
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Input: candidates = [2,3,5], target = 8,
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A solution set is:
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[
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[2,2,2,2],
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[2,3,3],
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[3,5]
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]
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## 题目大意
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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。
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candidates 中的数字可以无限制重复被选取。
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## 解题思路
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- 题目要求出总和为 sum 的所有组合，组合需要去重。
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- 这一题和第 47 题类似，只不过元素可以反复使用。
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## 代码
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```go
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package leetcode
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import "sort"
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func combinationSum(candidates []int, target int) [][]int {
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if len(candidates) == 0 {
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return [][]int{}
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}
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c, res := []int{}, [][]int{}
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sort.Ints(candidates)
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findcombinationSum(candidates, target, 0, c, &res)
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return res
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}
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func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
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if target <= 0 {
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if target == 0 {
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b := make([]int, len(c))
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copy(b, c)
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*res = append(*res, b)
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}
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return
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}
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for i := index; i < len(nums); i++ {
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if nums[i] > target { // 这里可以剪枝优化
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break
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}
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c = append(c, nums[i])
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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
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c = c[:len(c)-1]
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}
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}
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```
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