【leetcode】698. Partition to K Equal Sum Subsets 【leetcode】473. Matchsticks to Square 【leetcode】473. Matchsticks to Square

题目如下：

解题思路：本题是【leetcode】473. Matchsticks to Square的姊妹篇，唯一的区别是【leetcode】473. Matchsticks to Square指定了分成四个子数组，而本题分成的份数不定，作为参数输入。另外，本题的测试用例好像复杂一点，因此我过滤掉了nums中值等于avg的元素，不参与后面的DFS计算，提高效率。

代码如下：

class Solution(object):
    def canPartitionKSubsets(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        border = k
        if sum(nums) % border != 0:
            return False
        avg = sum(nums) / border
        #过滤掉了nums中值等于avg的元素，不参与后面的DFS计算
        newnums = []
        for i in nums:
            if i == avg:
                border -= 1
                continue
            newnums.append(i)
        nums = newnums[:]
        nums.sort()
        queue = [[x] for x in xrange(len(nums))]
        res = []
        visit = [0 for x in xrange(len(nums))]
        while len(queue) > 0:
            nl = queue.pop(0)
            amount = 0
            for i in nl:
                amount += nums[i]
            if amount == avg:
                res.append(nl)
                for i in nl:
                    visit[i] = 1
                continue
            tl = []
            for i in xrange(nl[-1] + 1, len(nums)):
                if amount + nums[i] <= avg:
                    tl = nl[:]
                    tl.append(i)
                    queue.append(tl)
        if len(res) < border:
            return False
        if sum(visit) != len(visit):
            return False
        queue = []
        for i in res:
            queue.append((set(i), 1))
        # print queue
        while len(queue) > 0:
            ns, count = queue.pop(0)
            if count == border and len(ns) == len(nums):
                # print ns
                return True
            for i in res:
                # print ns | set(i)
                if len(ns | set(i)) == len(ns) + len(i):
                    queue.insert(0, (ns | set(i), count + 1))

        return False