动态规划_leetcode343
Posted AceKo
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#coding=utf-8
# 递归
class Solution1(object):
def integerBreak(self, n):
"""
:type n: int
:rtype: int
"""
self.memo = [-1 for i in range(n+1)]
# 将n进行分割（至少分割两部分）,可以获得的最大乘积
def breakInterger(self,n):
if n == 1:
return 1
res = -1
# i + (n-i)
for i in range(1,n):
res = max(res, i* (n-i),self.breakInterger(i))
return res
# 记忆化递归
class Solution2(object):
def integerBreak(self, n):
"""
:type n: int
:rtype: int
"""
pass
# 将n进行分割（至少分割两部分）,可以获得的最大乘积
def breakInterger(self,n):
if n == 1:
return 1
if self.memo[n] != -1:
return self.memo[n]
res = -1
# i + (n-i)
for i in range(1,n):
res = max( res, i* (n-i),self.breakInterger(i))
self.memo[n] = res
return self.res
# 动态规划: 先解决最基本的问题，由底向上解决原问题
class Solution3(object):
def integerBreak(self, n):
"""
:type n: int
:rtype: int
"""
self.breakInterger(n)
# 将n进行分割（至少分割两部分）,可以获得的最大乘积
def breakInterger(self,n):
# 将n进行分割（至少分割两部分）,可以获得的最大乘积
memo = [-1 for i in range(n+1)]
memo[1] = 1
#memo[n]
# for i in range(1,n):
# memo[n] = max(memo[n],i * (n-i), i * memo[n-i])
# #memo[2]
# for i in range(1,2):
# memo[2] = max(memo[2],i * (2-i),i*memo[2-i])
#
#
# #memo[3]
# for i in range(1,3):
# memo[3] = max(memo[3],i * (3-i), i *memo[3-i])
#
# #memo[4]
# for i in range(1, 4):
# memo[4] = max(memo[4], i * (4 - i), i * memo[4 - i])
for i in range(1,n+1):
for j in range(1,i):
memo[i] = max(memo[i], j * (i-j) ,j * memo[i-j])
print memo[n]
return memo[n]
s = Solution3()
s.integerBreak(10)
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