Leetcode 279. 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。

 1 class Solution(object):
 2     def numSquares(self, n):
 3         """
 4         :type n: int
 5         :rtype: int
 6         """
 7         MAX = 2147483647
 8         m = 1
 9         perfect = [m]
10         while m**2 <= n:
11             perfect.append(m**2)
12             m += 1
13         
14         dp = [MAX]*(n+1)
15         dp[0] = 1
16         for x in perfect:
17             dp[x] = 1
18         
19         for i in range(2, n+1):
20             curP = [x for x in perfect if x <= i]
21             for j in curP:
22                 dp[i] = min(dp[j] + dp[i-j], dp[i])
23         
24         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。

 1 class Solution(object):
 2     def numSquares(self, n):
 3         """
 4         :type n: int
 5         :rtype: int
 6         """
 7         while n % 4 == 0:
 8             n = n//4
 9         
10         if n % 8 == 7:
11             return 4
12         
13         a = 0
14         while a**2 <= n:
15             b = int(math.sqrt(n - a**2))
16             if a**2 + b**2 == n:
17                 if a>0 and b>0:
18                     return 2
19                 if a == 0 and b>0:
20                     return 1
21                 if a>0 and b==0:
22                     return 1
23             a += 1
24         return 3

 

posted @ 2016-12-15 14:24  lettuan  阅读(179)  评论(0)  编辑  收藏  举报