标签：possible position || Plan ide class style false bsp
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 Explanation: It‘s possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.
Note:
1 <= k <= len(nums) <= 16.0 < nums[i] < 10000.Assume sum is the sum of nums[] . The dfs process is to find a subset of nums[] which sum equals to sum/k. We use an array visited[] to record which element in nums[] is used. Each time when we get a cur_sum = sum/k, we will start from position 0 in nums[] to look up the elements that are not used yet and find another cur_sum = sum/k.
1 class Solution { 2 public boolean canPartitionKSubsets(int[] nums, int k) { 3 int sum = 0; 4 for (int num : nums) sum += num; 5 if (k <= 0 || sum % k != 0) return false; 6 int[] visited = new int[nums.length]; 7 return canPartition(nums, visited, 0, k, 0, 0, sum / k); 8 } 9 10 public boolean canPartition(int[] nums, int[] visited, int start_index, int k, int cur_sum, int cur_num, int target) { 11 if (k == 1) return true; 12 if (cur_sum == target && cur_num > 0) return canPartition(nums, visited, 0, k - 1, 0, 0, target); 13 for (int i = start_index; i < nums.length; i++) { 14 if (visited[i] == 0) { 15 visited[i] = 1; 16 if (canPartition(nums, visited, i + 1, k, cur_sum + nums[i], cur_num++, target)) { 17 return true; 18 } 19 visited[i] = 0; 20 } 21 } 22 return false; 23 } 24 }
Partition to K Equal Sum Subsets
标签：possible position || Plan ide class style false bsp
原文地址：https://www.cnblogs.com/beiyeqingteng/p/10817356.html
