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),我们直接
classSolution {public:boolcanPartition(vector<int>&nums) { 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...
classSolution {publicbooleancanPartitionKSubsets(int[] nums,intk) {intsum = 0;for(intnum : nums) sum +=num;if(sum % k != 0)returnfalse;inttgt = sum /k;boolean[] visited =newboolean[nums.length];returndfs(nums, visited, 0, k, 0, tgt); }publicbooleandfs(int[] nums,boolean[] ...
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),...
1013. Partition Array Into Three Parts With Equal Sum(Leetcode每日一题-2020.03.11) Problem Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums. Formally, we can partition the array if we can find indexes i+1...
Partition Equal Subset Sum 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: E...leetcode 416. Partition Equal Subset Sum 题目: Given a non-empty ...
leetcode: https://leetcode.com/problems/partition-equal-subset-sum/description/ 写了之后,发现这题跟01背包还有点区别。但是写看这个吧。 暴力搜索的方法。就是每个取或者不去。 超时是必然的。接下来,我们把目前 考虑index和目标target记录下来: 这下通过了。但是我们发现这里 存在众多的递归调用。这就是因为...
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.
[LeetCode] 698. Partition to K Equal Sum Subsets Problem Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4...
public boolean canPartitionKSubsets(int[] nums, int k) { int sum = 0; for (int num: nums) sum += num; if (k == 0 || sum%k != 0 || sum < k) return false; int target = sum/k; return dfs(nums, new boolean[nums.length], 0, k, 0, target); ...