@@ -130,40 +130,93 @@ class Solution {
130130```
131131
132132Python:
133+ 133134``` Python
134135class Solution :
135136 def rob (self nums : List[int ]) -> int :
136- # 在198入门级的打家劫舍问题上分两种情况考虑
137- # 一是不偷第一间房,二是不偷最后一间房
138- if len (nums)== 1 : # 题目中提示nums.length>=1,所以不需要考虑len(nums)==0的情况
137+ if len (nums) == 0 :
138+ return 0
139+ if len (nums)== 1 :
139140 return nums[0 ]
140- val1= self .roblist(nums[1 :])# 不偷第一间房
141- val2= self .roblist(nums[:- 1 ])# 不偷最后一间房
142- return max (val1,val2)
143- 144- def roblist (self nums ):
145- l= len (nums)
146- dp= [0 ]* l
147- dp[0 ]= nums[0 ]
148- for i in range (1 ,l):
149- if i== 1 :
150- dp[i]= max (dp[i- 1 ],nums[i])
151- else :
152- dp[i]= max (dp[i- 1 ],dp[i- 2 ]+ nums[i])
153- return dp[- 1 ]
141+ 142+ result1 = self .robRange(nums, 0 , len (nums) - 2 ) # 情况二
143+ result2 = self .robRange(nums, 1 , len (nums) - 1 ) # 情况三
144+ return max (result1, result2)
145+ # 198.打家劫舍的逻辑
146+ def robRange (self nums : List[int ], start : int , end : int ) -> int :
147+ if end == start:
148+ return nums[start]
149+ 150+ prev_max = nums[start]
151+ curr_max = max (nums[start], nums[start + 1 ])
152+ 153+ for i in range (start + 2 , end + 1 ):
154+ temp = curr_max
155+ curr_max = max (prev_max + nums[i], curr_max)
156+ prev_max = temp
157+ 158+ return curr_max
159+ 154160```
161+ 2维DP
155162``` python
156- class Solution :# 二维dp数组写法
163+ class Solution :
157164 def rob (self nums : List[int ]) -> int :
158- if len (nums)< 3 : return max (nums)
159- return max (self .default(nums[:- 1 ]),self .default(nums[1 :]))
160- def default (self nums ):
161- dp = [[0 ,0 ] for _ in range (len (nums))]
165+ if len (nums) < 3 :
166+ return max (nums)
167+ 168+ # 情况二:不抢劫第一个房屋
169+ result1 = self .robRange(nums[:- 1 ])
170+ 171+ # 情况三:不抢劫最后一个房屋
172+ result2 = self .robRange(nums[1 :])
173+ 174+ return max (result1, result2)
175+ 176+ def robRange (self nums ):
177+ dp = [[0 , 0 ] for _ in range (len (nums))]
162178 dp[0 ][1 ] = nums[0 ]
163- for i in range (1 ,len (nums)):
164- dp[i][0 ] = max (dp[i- 1 ])
165- dp[i][1 ] = dp[i- 1 ][0 ] + nums[i]
179+ 180+ for i in range (1 , len (nums)):
181+ dp[i][0 ] = max (dp[i - 1 ])
182+ dp[i][1 ] = dp[i - 1 ][0 ] + nums[i]
183+ 166184 return max (dp[- 1 ])
185+ 186+ 187+ 188+ ```
189+ 190+ 优化版
191+ ``` python
192+ class Solution :
193+ def rob (self nums : List[int ]) -> int :
194+ if not nums: # 如果没有房屋,返回0
195+ return 0
196+ 197+ if len (nums) == 1 : # 如果只有一个房屋,返回该房屋的金额
198+ return nums[0 ]
199+ 200+ # 情况二:不抢劫第一个房屋
201+ prev_max = 0 # 上一个房屋的最大金额
202+ curr_max = 0 # 当前房屋的最大金额
203+ for num in nums[1 :]:
204+ temp = curr_max # 临时变量保存当前房屋的最大金额
205+ curr_max = max (prev_max + num, curr_max) # 更新当前房屋的最大金额
206+ prev_max = temp # 更新上一个房屋的最大金额
207+ result1 = curr_max
208+ 209+ # 情况三:不抢劫最后一个房屋
210+ prev_max = 0 # 上一个房屋的最大金额
211+ curr_max = 0 # 当前房屋的最大金额
212+ for num in nums[:- 1 ]:
213+ temp = curr_max # 临时变量保存当前房屋的最大金额
214+ curr_max = max (prev_max + num, curr_max) # 更新当前房屋的最大金额
215+ prev_max = temp # 更新上一个房屋的最大金额
216+ result2 = curr_max
217+ 218+ return max (result1, result2)
219+ 167220
168221```
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