leetcode279 Perfect Squares

Posted 2021-03-07 yawenw

 1 """
 2 Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
 3 Example 1:
 4 Input: n = 12
 5 Output: 3
 6 Explanation: 12 = 4 + 4 + 4.
 7 Example 2:
 8 Input: n = 13
 9 Output: 2
10 Explanation: 13 = 4 + 9.
11 """
12 """
13 dp[0] = 0
14 dp[1] = dp[0]+1 = 1
15 dp[2] = dp[1]+1 = 2
16 dp[3] = dp[2]+1 = 3
17 dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 }
18       = Min{ dp[3]+1, dp[0]+1 }
19       = 1
20 dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 }
21       = Min{ dp[4]+1, dp[1]+1 }
22       = 2
23                         .
24                         .
25                         .
26 dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 }
27        = Min{ dp[12]+1, dp[9]+1, dp[4]+1 }
28        = 2
29                         .
30                         .
31                         .
32 dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1
33 dp[n] 表示以n为和的最少平方的和的个数（所求）。
34 dp 数组所有下标已经为完全平方数的数（如1,4,9...）置为 1，加一层 j 遍历找到当前 i 下长度最小的组合。
35 动态方程的意思是：对于每个 i ，比 i 小一个完全平方数的那些数中最小的个数+1就是所求，也就是 dp [ i - j * j ] + 1 。
36 """
37 class Solution1:
38     def numSquares(self, n):
39         dp = [float(‘inf‘)]*(n+1)
40         i = 1
41         while i*i <= n:
42             dp[i*i] = 1
43             i += 1
44         for i in range(1, n+1):
45             j = 1
46             while j*j <= i:
47                 dp[i] = min(dp[i], dp[i-j*j]+1)
48                 j += 1
49         return dp[n]