LeetCode 473 - Matchsticks to Square - Medium (Python)

Posted 2021-03-13 sky37

Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:

Input: [1,1,2,2,2]
Output: true

Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.

 

Example 2:

Input: [3,3,3,3,4]
Output: false

Explanation: You cannot find a way to form a square with all the matchsticks.

卖火柴的小女孩。需要注意的是全部的火柴都要用完。以及不要有一种思维定式，火柴不一定是要连着拿的。比如[5,5,5,5,4,4,4,4,3,3,3,3] 不连续取，是可能形成一个square的。但是如果思维定式火柴连着取，那么得到
的结果会是错的。

错误写法

class Solution:
    def makesquare(self, nums: List[int]) -> bool:
        if not nums or len(nums) == 0:
            return False 
        if sum(nums) % 4 != 0:
            return False 
        
        length = sum(nums) // 4 
        if any (num > length for num in nums):
            return False 
        nums.sort()
        
        return self.dfs(nums, 0, length, 0, 0)
    
    
    def dfs(self, nums, start, length, temp, count):
        if count == 4: return True 
        if start > len(nums): 
            return True 
        if temp == length:
            return self.dfs(nums, start, length, 0, count+1)
        
        for i in range(start, len(nums)):
            if temp+nums[i] > length:
                break 
            if self.dfs(nums, i+1, length, temp+nums[i], count): # wrong！！！
                return True 

正确的思考方式是用dfs遍历可以得到的所有火柴的摆放顺序。如果一味的遍历，那么最后肯定会TLE。一个比较好的方式是构建一个target = [l] * 4. l是可以摆成的火柴的边长。 如果我们可以把这个target的值都变为0，

那么意味着是可以摆成正方形的。

步骤：1 构建target

         2 dfs 函数 def dfs(nums, target, pos). dfs跳出条件是 pos == len(nums). 对target进行从0到4的遍历（不包含4），每一次遍历的过程中，对nums也进行一次遍历，如果target[i] >= num, 意味着这个num可能成为

这条边长的candidate，那么我们在这个基础上进行下一层的dfs。注意要把target变为减掉num的target。self.dfs(nums, pos+1, target[:i] + [target[i]-nums[pos]]+target[i+1:])。如果下一层的dfs返回值是True，意味着这些火柴是可以摆放成功的。

比较疑惑的是当nums从大到小排的时候，不会超时。但是从小到大排，就会超。

class Solution:
    def makesquare(self, nums: List[int]) -> bool:
        if not nums or len(nums) == 0:
            return False 
        if sum(nums) % 4 != 0:
            return False 
        
        length = sum(nums) // 4 
        if any (num > length for num in nums):
            return False 
        nums.sort(reverse = True)
        target = [length] * 4 
        return self.dfs(nums, 0,target)
    
    
    def dfs(self, nums, pos, target):
        if pos == len(nums):
            return True 
        for i in range(4):
            if target[i] >= nums[pos]: 
                if self.dfs(nums, pos+1, target[:i] + [target[i]-nums[pos]]+target[i+1:]):
                    return True 
             
        return False