【leetCode】House Robber II python實現
阿新 • • 發佈:2019-01-11
原題連結
實現原理解析
該題在House Robber的基礎上讓首位連結形成環,那麼即表示第一個和最後一個不能同時被搶,則問題分解為House Robber(nums[0:len(nums)-1])和House Robber(nums[1:len(nums)]),兩者中比較大的那個即為結果
python程式碼實現
class Solution(object):
def rob(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
length = len(nums)
if length == 0:
return 0
elif length == 1:
return nums[0]
elif length ==2:
return max(nums[0], nums[1])
return max(self.dfs(nums[1:]), self.dfs(nums[:-1]))
def dfs(self, nums):
if len(nums) == 2:
return max(nums[0], nums[1 ])
res = [0 for each in nums]
res[0] = nums[0]
res[1] = max(nums[0], nums[1])
for index in xrange(2,len(nums)):
res[index] = max(res[index -2] + nums[index], res[index - 1])
return res[len(nums) - 1]