更新0213.打家劫舍II.md 更简洁的python 版本

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Verolilo
2022-02-16 10:22:19 +08:00
committed by GitHub
parent 2716470e63
commit 6d5a0d0340

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@ -123,22 +123,24 @@ Python
```Python
class Solution:
def rob(self, nums: List[int]) -> int:
if (n := len(nums)) == 0:
return 0
if n == 1:
return nums[0]
result1 = self.robRange(nums, 0, n - 2)
result2 = self.robRange(nums, 1, n - 1)
return max(result1 , result2)
#在198入门级的打家劫舍问题上分两种情况考虑
#一是不偷第一间房,二是不偷最后一间房
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
return nums[0]
val1=self.roblist(nums[1:])#不偷第一间房
val2=self.roblist(nums[:-1])#不偷最后一间房
return max(val1,val2)
def robRange(self, nums: List[int], start: int, end: int) -> int:
if end == start: return nums[start]
dp = [0] * len(nums)
dp[start] = nums[start]
dp[start + 1] = max(nums[start], nums[start + 1])
for i in range(start + 2, end + 1):
dp[i] = max(dp[i -2] + nums[i], dp[i - 1])
return dp[end]
def robRange(self,nums):
l=len(nums)
dp=[0]*l
dp[0]=nums[0]
for i in range(1,l):
if i==1:
dp[i]=max(dp[i-1],nums[i])
else:
dp[i]=max(dp[i-1],dp[i-2]+nums[i])
return dp[-1]
```
javascipt: