181 lines
4.9 KiB
Markdown
181 lines
4.9 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/combination-sum/description/
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## 题目描述
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```
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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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```
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## 思路
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这道题目是求集合，并不是`求极值`，因此动态规划不是特别切合，因此我们需要考虑别的方法。
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这种题目其实有一个通用的解法，就是回溯法。
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网上也有大神给出了这种回溯法解题的
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[通用写法](https://leetcode.com/problems/combination-sum/discuss/16502/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning))，这里的所有的解法使用通用方法解答。
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除了这道题目还有很多其他题目可以用这种通用解法，具体的题目见后方相关题目部分。
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我们先来看下通用解法的解题思路，我画了一张图：
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通用写法的具体代码见下方代码区。
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## 关键点解析
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- 回溯法
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- backtrack 解题公式
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## 代码
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* 语言支持: Javascript，Python3
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```js
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/*
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* @lc app=leetcode id=39 lang=javascript
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*
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* [39] Combination Sum
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*
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* https://leetcode.com/problems/combination-sum/description/
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*
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* algorithms
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* Medium (46.89%)
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* Total Accepted: 326.7K
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* Total Submissions: 684.2K
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* Testcase Example: '[2,3,6,7]\n7'
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*
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* Given a set of candidate numbers (candidates) (without duplicates) and a
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* target number (target), find all unique combinations in candidates where the
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* candidate numbers sums to target.
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*
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* The same repeated number may be chosen from candidates unlimited number of
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* times.
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*
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* Note:
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*
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*
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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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*
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*
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* Example 1:
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*
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*
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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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*
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*
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* Example 2:
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*
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*
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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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*
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*/
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function backtrack(list, tempList, nums, remain, start) {
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if (remain < 0) return;
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else if (remain === 0) return list.push([...tempList]);
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for (let i = start; i < nums.length; i++) {
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tempList.push(nums[i]);
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backtrack(list, tempList, nums, remain - nums[i], i); // 数字可以重复使用， i + 1代表不可以重复利用
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tempList.pop();
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}
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}
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/**
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* @param {number[]} candidates
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* @param {number} target
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* @return {number[][]}
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*/
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var combinationSum = function(candidates, target) {
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const list = [];
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backtrack(list, [], candidates.sort((a, b) => a - b), target, 0);
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return list;
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};
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```
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Python3 Code:
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```python
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class Solution:
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def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
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"""
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回溯法，层层递减，得到符合条件的路径就加入结果集中，超出则剪枝；
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主要是要注意一些细节，避免重复等；
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"""
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size = len(candidates)
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if size <= 0:
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return []
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# 先排序，便于后面剪枝
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candidates.sort()
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path = []
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res = []
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self._find_path(target, path, res, candidates, 0, size)
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return res
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def _find_path(self, target, path, res, candidates, begin, size):
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"""沿着路径往下走"""
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if target == 0:
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res.append(path.copy())
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else:
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for i in range(begin, size):
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left_num = target - candidates[i]
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# 如果剩余值为负数，说明超过了，剪枝
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if left_num < 0:
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break
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# 否则把当前值加入路径
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path.append(candidates[i])
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# 为避免重复解，我们把比当前值小的参数也从下一次寻找中剔除，
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# 因为根据他们得出的解一定在之前就找到过了
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self._find_path(left_num, path, res, candidates, i, size)
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# 记得把当前值移出路径，才能进入下一个值的路径
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path.pop()
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```
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## 相关题目
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- [40.combination-sum-ii](./40.combination-sum-ii.md)
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- [46.permutations](./46.permutations.md)
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- [47.permutations-ii](./47.permutations-ii.md)
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- [78.subsets](./78.subsets.md)
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- [90.subsets-ii](./90.subsets-ii.md)
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- [113.path-sum-ii](./113.path-sum-ii.md)
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- [131.palindrome-partitioning](./131.palindrome-partitioning.md)
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