【leetcode】473. Matchsticks to Square

题目如下：

解题思路：居然把卖火柴的小女孩都搬出来了。题目的意思是输入一个数组，判断能否把数组分成四个子数组，使得每个子数组的和相等。首先我们可以很容易的求出每个子数组的和应该是avg = sum(nums)/4，接下来我的思路是求出nums中所有和等于avg的子数组，子数组中保存元素下标，然后找出四个不存在相同下标的子数组求并集，如果并集长度刚好等于len(nums),那么就是符合条件的，返回true。如果在子数组列表中找到四个符合条件的子数组，我用的是DFS，为什么不用BFS？因为题目要求是找到一组符合条件的子数组即可，DFS显然要比BFS快。 

代码如下：

class Solution(object):
    def makesquare(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        nums.sort()
        border = 4
        if sum(nums)%border != 0 :
            return False
        avg = sum(nums)/border
        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