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LeetCode题解39.combination-sum程序员分享

本文介绍了LeetCode题解39.combination-sum程序员分享,有助于帮助完成毕业设计以及求职,是一篇很好的资料。

对技术面试,学习经验等有一些体会,在此分享。

题目地址

https://leetcode.com/problems/combination-sum/description/

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] ]  

思路

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。

这种题目其实有一个通用的解法,就是回溯法。
网上也有大神给出了这种回溯法解题的
通用写法,这里的所有的解法使用通用方法解答。
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。

我们先来看下通用解法的解题思路,我画了一张图:

LeetCode题解39.combination-sum

通用写法的具体代码见下方代码区。

关键点解析

  • 回溯法
  • backtrack 解题公式

代码

  • 语言支持: Javascript,Python3
/*  * @lc app=leetcode id=39 lang=javascript  *  * [39] Combination Sum  *  * https://leetcode.com/problems/combination-sum/description/  *  * algorithms  * Medium (46.89%)  * Total Accepted:    326.7K  * Total Submissions: 684.2K  * Testcase Example:  '[2,3,6,7]n7'  *  * 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]  * ]  *  */  function backtrack(list, tempList, nums, remain, start) {   if (remain < 0) return;   else if (remain === 0) return list.push([...tempList]);   for (let i = start; i < nums.length; i++) {     tempList.push(nums[i]);     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     tempList.pop();   } } /**  * @param {number[]} candidates  * @param {number} target  * @return {number[][]}  */ var combinationSum = function(candidates, target) {   const list = [];   backtrack(list, [], candidates.sort((a, b) => a - b), target, 0);   return list; };

Python3 Code:

class Solution:     def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:         """         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;         主要是要注意一些细节,避免重复等;         """         size = len(candidates)         if size <= 0:             return []                  # 先排序,便于后面剪枝         candidates.sort()                  path = []         res = []         self._find_path(target, path, res, candidates, 0, size)                  return res              def _find_path(self, target, path, res, candidates, begin, size):         """沿着路径往下走"""         if target == 0:             res.append(path.copy())         else:             for i in range(begin, size):                 left_num = target - candidates[i]                 # 如果剩余值为负数,说明超过了,剪枝                 if left_num < 0:                     break                 # 否则把当前值加入路径                 path.append(candidates[i])                 # 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,                 # 因为根据他们得出的解一定在之前就找到过了                 self._find_path(left_num, path, res, candidates, i, size)                 # 记得把当前值移出路径,才能进入下一个值的路径                 path.pop()

相关题目

  • 40.combination-sum-ii
  • 46.permutations
  • 47.permutations-ii
  • 78.subsets
  • 90.subsets-ii
  • 113.path-sum-ii
  • 131.palindrome-partitioning

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