# -*- 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
# 从memory数组中拿到已经计算完的
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] 表示将数字i分割(至少分割为2份)后最大的成绩。
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)
leetcode——integer break(343)
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