[一起来刷leetcode吧][2]--No.698 Partition to K Equal Sum Subsets.md

本文介绍LeetCode上第698题Partition to K Equal Sum Subsets的解决思路及代码实现。该题旨在判断是否能将一个整数数组分割成k个非空子集且各子集之和相等。文章首先概述了问题背景，随后详细阐述了解决方案，包括排序和递归搜索等关键步骤。

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这是leetcode的第698题--Partition to K Equal Sum Subsets

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.

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

思路先排好序，然后对当前剩余的最大的数，判断是否存在一组数使等于平均值。通过将两个数相加，一直归并，如果不成立，则回溯，直到成立，或者没有选择则返回False

show me the code

class Solution:
    def canPartitionKSubsets(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        n=len(nums)
        s , res= divmod(sum(nums),k)
        if res!=0 :return False
        nums.sort()
        return self._util(nums,s,k)

    def _util(self,nums,s,n):
        #print(nums,s,n)
        if nums == [] and n==0:return True
        elif nums ==[] or n==0:return False
        if nums[-1] >s:return False
        elif nums[-1] ==s:return self._util(nums[:-1],s,n-1)
        else:
            if nums[-1] nums[0]>s:return False
            elif nums[-1] nums[0]==s:return self._util(nums[1:-1],s,n-1)
            else:
                ln = len(nums)
                i=1
                for i in range(1,ln-1):
                    if nums[i] nums[-1]==s:
                        nums.remove(nums[i])
                        return self._util(nums[:-1],s,n-1)
                    elif nums[i] nums[-1]>s:
                        break
                for j in range(i):
                    li =nums[0:j]  nums[j 1:]
                    li[-1] =nums[j]
                    if self._util(li,s,n):return True
                else:return False