首先一定不能构成两个和相等的情况可以直接排除,对于一个数组 num,我们记 num 的两个子数组的和为 x,有 x + x = sum(num),所以要求 num 的和一定是偶数,那么如果整个 num 的和是奇数,我们直接返回 False;子数组的和为 x,那么数组 num 的最大值一定小于等于 x,如果最大值大于 x(即 num 和的一半),...
bitset<5001> bits(1);intsum = accumulate(nums.begin(), nums.end(),0);for(intnum : nums) bits |= bits <<num;return(sum %2==0) && bits[sum >>1]; } }; Github 同步地址: https://github.com/grandyang/leetcode/issues/416 类似题目: Partition to K Equal Sum Subset 参考资料: http...
problem:https://leetcode.com/problems/partition-equal-subset-sum/ 经典背包问题。找到是否存在恰好装满sum / 2的物体,可以优化为1D的。 classSolution {public:boolcanPartition(vector<int>&nums) {intn =nums.size();intsum = accumulate(nums.begin(), nums.end(),0);if(sum %2)returnfalse; sum/=...
给你一个只包含正整数的非空数组nums。请你判断是否可以将这个数组分割成两个子集,使得两个子集的元素和相等。 示例1: 输入:nums = [1,5,11,5]输出:true解释:数组可以分割成 [1, 5, 5] 和 [11] 。 示例2: 输入:nums = [1,2,3,5]输出:false解释:数组不能分割成两个元素和相等的子集。 提示: ...
本次題目也是關於動態規劃的練習,Partition Equal Subset Sum,題目要求如下: Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of ... 查看原文 [leetcode]416. Partition Equal Subset Sum 难0.0] Given a non-empty...
Problem Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal. Note: Each of the array element will not exceed 100.
Java | LeetCode | 416. Partition Equal Subset Sum | 背包问题 | 动态规划 416. Partition Equal Subset Sum | 背包问题 | 动态规划 问题描述 Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in bot...
Output:falseExplanation: The array cannot be partitioned into equal sum subsets. https://leetcode.com/problems/partition-equal-subset-sum/discuss/90592/01-knapsack-detailed-explanationNice solution. I just figured out the differences betweenthisproblem and 518. Coin Change 2. Inthisproblem, we CANN...
Breadcrumbs leetcode-solutions /cpp / 0416-partition-equal-subset-sum.cpp Latest commit Cannot retrieve latest commit at this time. HistoryHistory File metadata and controls Code Blame 42 lines (36 loc) · 1.12 KB Raw /* Given non-empty, non-negative integer array nums, find if: Can be...
递归函数:基于已选的元素(和为curSum),从i开始继续选,能否选出和为sum/2的子集。每次递归,都有两个选择:选nums[i]。基于选它,往下继续选(递归):dfs(curSum + nums[i], i + 1) 不选nums[i]。基于不选它,往下继续选(递归):dfs(curSum, i + 1)...