leetcode279 Perfect Squares

"""
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
Example 1:
Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
Example 2:
Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.
"""
"""
dp[0] = 0
dp[1] = dp[0]+1 = 1
dp[2] = dp[1]+1 = 2
dp[3] = dp[2]+1 = 3
dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 }
      = Min{ dp[3]+1, dp[0]+1 }
      = 1
dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 }
      = Min{ dp[4]+1, dp[1]+1 }
      = 2
                        .
                        .
                        .
dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 }
       = Min{ dp[12]+1, dp[9]+1, dp[4]+1 }
       = 2
                        .
                        .
                        .
dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1
dp[n] 表示以n为和的最少平方的和的个数（所求）。
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。
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 。
"""
class Solution1:
    def numSquares(self, n):
        dp = [float('inf')]*(n+1)
        i = 1
        while i*i <= n:
            dp[i*i] = 1
            i += 1
        for i in range(1, n+1):
            j = 1
            while j*j <= i:
                dp[i] = min(dp[i], dp[i-j*j]+1)
                j += 1
        return dp[n]