Leetcode Perfect Square

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

这道题首先想到的算法是DP。每个perfect square number对应的解都是1。先生成一个n+1长的DP list。对于每个i，可以用dp[i] = min(dp[j] + dp[i-j], dp[i])来更新，这里j 是<= i 的一个perfect square number。但是DP的算法超时。

 class Solution(object):
     def numSquares(self, n):
         """
         :type n: int
         :rtype: int
         """
         MAX = 2147483647
         m = 1
         perfect = [m]
         while m**2 <= n:
             perfect.append(m**2)
             m += 1

         dp = [MAX]*(n+1)
         dp[0] = 1
         for x in perfect:
             dp[x] = 1

         for i in range(2, n+1):
             curP = [x for x in perfect if x <= i]
             for j in curP:
                 dp[i] = min(dp[j] + dp[i-j], dp[i])

         return dp[-1]

解法二： 来自 https://www.cnblogs.com/grandyang/p/4800552.html

任何一个正整数都可以写成最多4个数的平方和。然后又两条性质可以简化题目：

1. 4q 和 q 的答案一样，i.e. 3, 12。

2. 如果一个数除以8余7 <=> 答案是4。

 class Solution(object):
     def numSquares(self, n):
         """
         :type n: int
         :rtype: int
         """
         while n % 4 == 0:
             n = n//4

         if n % 8 == 7:
             return 4

         a = 0
         while a**2 <= n:
             b = int(math.sqrt(n - a**2))
             if a**2 + b**2 == n:
                 if a>0 and b>0:
                     return 2
                 if a == 0 and b>0:
                     return 1
                 if a>0 and b==0:
                     return 1
             a += 1
         return 3