183 lines
5.1 KiB
Markdown
183 lines
5.1 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/combination-sum-ii/description/
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## 题目描述
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```
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Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
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Each number in candidates may only be used once in the combination.
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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 = [10,1,2,7,6,1,5], target = 8,
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A solution set is:
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[
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[1, 7],
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[1, 2, 5],
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[2, 6],
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[1, 1, 6]
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]
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Example 2:
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Input: candidates = [2,5,2,1,2], target = 5,
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A solution set is:
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[
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[1,2,2],
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[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=40 lang=javascript
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*
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* [40] Combination Sum II
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*
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* https://leetcode.com/problems/combination-sum-ii/description/
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*
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* algorithms
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* Medium (40.31%)
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* Total Accepted: 212.8K
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* Total Submissions: 519K
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* Testcase Example: '[10,1,2,7,6,1,5]\n8'
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*
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* Given a collection of candidate numbers (candidates) and a target number
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* (target), find all unique combinations in candidates where the candidate
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* numbers sums to target.
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*
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* Each number in candidates may only be used once in the combination.
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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 = [10,1,2,7,6,1,5], target = 8,
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* A solution set is:
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* [
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* [1, 7],
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* [1, 2, 5],
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* [2, 6],
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* [1, 1, 6]
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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 = [2,5,2,1,2], target = 5,
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* A solution set is:
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* [
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* [1,2,2],
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* [5]
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* ]
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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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// 和39.combination-sum 的其中一个区别就是这道题candidates可能有重复
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// 代码表示就是下面这一行
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if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
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tempList.push(nums[i]);
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backtrack(list, tempList, nums, remain - nums[i], i + 1); // i + 1代表不可以重复利用， i 代表数字可以重复使用
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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 combinationSum2 = 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 combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
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"""
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与39题的区别是不能重用元素，而元素可能有重复；
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不能重用好解决，回溯的index往下一个就行；
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元素可能有重复，就让结果的去重麻烦一些；
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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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# 还是先排序，主要是方便去重
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candidates.sort()
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path = []
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res = []
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self._find_path(candidates, path, res, target, 0, size)
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return res
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def _find_path(self, candidates, path, res, target, begin, size):
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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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if left_num < 0:
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break
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# 如果存在重复的元素，前一个元素已经遍历了后一个元素与之后元素组合的所有可能
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if i > begin and candidates[i] == candidates[i-1]:
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continue
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path.append(candidates[i])
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# 开始的 index 往后移了一格
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self._find_path(candidates, path, res, left_num, i+1, size)
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path.pop()
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```
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## 相关题目
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- [39.combination-sum](./39.combination-sum.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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