leetcode——integer break(343)

# -*- encoding: utf-8 -*-
"""
Given a positive integer n, break it into the sum of
at least two positive integers and maximize the product of those integers.
Return the maximum product you can get.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.
Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.
Note: You may assume that n is not less than 2 and not larger than 58.
"""


class Solution:

    def integerBreak(self, n: int) -> int:
        if n == 1:
            return 1
        rest = -1
        for i in range(1, n):  # 1~ (n-1)
            rest = max(i * (n - i), i * self.integerBreak(n - i), rest)
        return rest

    def integerBreak2(self, n: int) -> int:
        self.memory = [-1] * (n + 1)
        return self.int_break(n)

    def int_break(self, n) -> int:
        if n == 1:
            return 1
        # 從memory數組中拿到已經計算完的
        if self.memory[n] != -1:
            return self.memory[n]

        rest = -1
        for i in range(1, n):
            rest = max(i * (n - i), i * self.int_break(n - i), rest)
        self.memory[n] = rest
        return self.memory[n]

    def integerBreak3(self, n: int) -> int:
        # memo[i] 表示將數字i分割（至少分割爲2份）後最大的成績。
        memo = [-1] * (n + 1)
        memo[1] = 1
        for i in range(2, n + 1):  # 2到n
            # 求解 memo[i]
            for j in range(1, i):
                memo[i] = max(j * (i - j), j * memo[i - j], memo[i])
        return memo[n]


if __name__ == '__main__':
    solution = Solution()
    result = solution.integerBreak3(100)  #
    print(result)