LeetCode.698 Partition to K Equal Sum Subsets（均分成k组和相等）

1.题目

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.

2.输入/输出样例

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.

3.参考代码

class Solution {
     
     public static boolean canPartitionKSubsets(int[] nums, int k) {
     
        // 思路：典型的背包问题，但是每个数据仅使用一次，保证每个装包之和相等。
        // 可对需要装包的数据排序，每次判断装包数据是否满足target（即平均值）,来决定是否可装下去。

        if (nums == null || nums.length == 0) {
     
            return false;
        }

        if (k == 1) {
     
            return true;
        }

        int sum = 0;
        for (int num : nums) {
     
            sum += num;
        }

        if (sum % k != 0) {
     
            return false;
        }

        boolean[] visited = new boolean[nums.length];
        Arrays.sort(nums);
        return dfs(nums, 0, nums.length - 1, visited, sum / k, k);
    }

    public static boolean dfs(int[] nums, int curSum, int startIndex, boolean[] visited, int target, int k) {
     
        if(k==0){
     
            return true;
        }
        // 满足当前curSum为target，且后续的剩下的数据能继续均分成k-1份
        if(curSum==target&&dfs(nums,0,nums.length-1,visited,target,k-1)){
     
            return true;
        }
        
        for(int i=startIndex;i>=0;--i){
     
            if(!visited[i]&&curSum+nums[i]<=target){
     
                visited[i]=true;
                // 接着查找后面的数据，满足k次平分
                if(dfs(nums,curSum+nums[i],i-1,visited,target,k)){
     
                    return true;
                }
                visited[i]=false;
            }
        }
        return false;
    }
}