algorithm - 求和为 n 的最少完全平方数

Question

问题是:

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

示例

给定 n = 12，返回 3 因为 12 = 4 + 4 + 4；给定 n = 13，返回 2，因为 13 = 4 + 9。

注意

The approach i have taken is similar to an Integer Knapsack problem with duplicates allowed. First i calculated all the perfect squares which are less than equal to the number n. Now once i have them, the problem is similar to the Integer Knapsack problem. I have a number n and a list of numbers. i want to choose minimum number of numbers from the list such that their sum is equal to n. This problem has a DP solution which i have used.

Out of 586 test cases, i passed 562 test cases and got TLE on the next one. The value of n for that testcase is 3428.

我提交的解决方案:

class Solution(object):
    def numSquares(self,n):
        if(n == 0):
            return 0
        if(n == 1):
            return 1
        squares = self.findSquares(n) # returns a list of perfect squares <= n
        rows = len(squares)
        cols = n + 1
        mat = []
        for i in range(rows):
            mat.append([0] * cols)

        for i in range(cols):
            mat[0][i] = i

        for i in range(rows):
            mat[i][0] = 0

        min = mat[0][cols - 1]
        for i in range(1,rows):
            for j in range(1,cols):
                if(j < squares[i]):
                    mat[i][j] = mat[i - 1][j]

                else:
                    mat[i][j] = self.min(mat[i - 1][j], (j // squares[i] + (mat[i][j % squares[i]])))
                if(j == cols - 1 and mat[i][j] < min):
                    min = mat[i][j]
        '''
        for i in range(rows):
            for j in range(cols):
                print(mat[i][j],end=" ")
            print()
            '''
        return min

    def min(self,a,b):
        if(a <= b):
            return a
        else:
            return b


    def findSquares(self,n):

        i = 1
        squares = []
        while (i * i <= n):
            squares.append(i * i)
            i = i + 1
        return squares
'''
x = Solution()

print(x.numSquares(43))
'''

提前致谢。

Answer 1

您可以将您的解决方案简化为:

def numSquares(self,n):
    if(n == 0):
        return 0
    if(n == 1):
        return 1
    squares = self.findSquares(n)
    rows = len(squares)
    cols = n + 1
    mat = [n] * cols
    mat[0] = 0

    for s in squares:
        for j in range(s,cols):
            mat[j] = min(mat[j], 1 + mat[j - s])

    return mat[n]

这避免了使用:

1. self.min 函数
2. 内循环中的除法/取模运算。
3. 二维数组

而且速度大约是原来的两倍。