classSolution {public:boolcanPartitionKSubsets(vector<int>& nums,intk) {intsize = nums.size();if(size ==0)returnfalse;if(k ==1)returntrue;if(k <1|| k >16|| k > size)returnfalse;intsum =0;for(inti:nums) { sum += i; }if(sum%k !=0)returnfalse;vector<bool>visited(size,fa...
classSolution {publicbooleancanPartitionKSubsets(int[] nums,intk) {if(nums ==null|| nums.length == 0) {returnfalse; }intsum = 0;for(inti = 0; i < nums.length; i++) { sum+=nums[i]; }if(sum % k != 0) {returnfalse; }inttarget = sum /k;boolean[] isUsed =newboolean[num...
classSolution{publicbooleancanPartitionKSubsets(int[] nums,intk){if(k ==1) {returntrue; }if(nums ==null|| nums.length ==0|| k > nums.length) {returnfalse; }intsum=0;intmax=0;for(inti:nums) { sum = sum + i ; max = Math.max(max,i); }if(sum % k !=0|| sum / k < ...
此时有sum(target) >= 0。又已知所有数字恰好用完,所以恰好有sum(target) = 0。故,当所有数字恰好用完时,target的每个位置都是0. Python代码: class Solution: def canPartitionKSubsets(self, nums, k): """ :type nums: List[int] :type k: int :rtype: bool """ if not nums or len(nums) <...
https://leetcode.com/problems/partition-to-k-equal-sum-subsets/description/ 题目: Given an array of integersnumsand a positive integerk, find whether it's possible to divide this array intoknon-empty subsets whose sums are all equal.
Leetcode 560[medium]. Subarray Sum Equals K 难度:medium Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Note: The length of the array is in range [... ...
1 <= k <= len(nums) <= 16. 0 < nums[i] < 10000. 这道题给了我们一个数组nums和一个数字k,问我们该数字能不能分成k个非空子集合,使得每个子集合的和相同。给了k的范围是[1,16],而且数组中的数字都是正数。这跟之前那道Partition Equal Subset Sum很类似,但是那道题只让分成两个子集合,所以问...
LeetCode 698. Partition to K Equal Sum Subsets 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 Output: True...
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); } private boolean dfs(int[] nums, boolean[] used, int start, int k, int sum, int target) { ...
Constraints:1<=N<=501<=S[i]<=1000SumofelementsofS willnotbe greater than1000.Example:Input:523416Output:8{Explanation:Thetwo subsetswithmaximum sum are[1,3,4]and[2,6]} I found this question in one of the interview challenge. I used the traditional sum of subsets logic to find all the...