目录
学习目标
学习内容
198.打家劫舍
213.打家劫舍II
337.打家劫舍III
学习目标
- 198.打家劫舍
- 213.打家劫舍II
- 337.打家劫舍III
学习内容
198.打家劫舍
198. 打家劫舍 - 力扣(LeetCode)https://leetcode.cn/problems/house-robber/
class Solution:
def rob(self, nums: List[int]) -> int:
n = len(nums)
dp = [0]*(n+1)
dp[1] = nums[0]
for i in range(2,n+1):
dp[i] = max(dp[i-1],dp[i-2]+nums[i-1])
return dp[n]
213.打家劫舍II
213. 打家劫舍 II - 力扣(LeetCode)https://leetcode.cn/problems/house-robber-ii/
class Solution:
def rob(self, nums: List[int]) -> int:
n = len(nums)
if n==1:return nums[0]
@cache
def dfs(i,first):
if i<0:return 0
if i==0:
if first:return nums[i]
return 0
return max(dfs(i-1,first),dfs(i-2,first)+nums[i])
return max(dfs(n-1,False),dfs(n-2,True))
class Solution:
def rob(self, nums: List[int]) -> int:
n = len(nums)
if n==1:return nums[0]
dp = [0]*(n+1)
dp[1] = nums[0] # 选首家不选尾家
for i in range(2,n):
dp[i] = max(dp[i-1],dp[i-2]+nums[i-1])
a = dp[n-1]
dp = [0]*(n+1)
for i in range(2,n+1):
dp[i] = max(dp[i-1],dp[i-2]+nums[i-1])
b = dp[n]
return max(a,b)
337.打家劫舍III
337. 打家劫舍 III - 力扣(LeetCode)https://leetcode.cn/problems/house-robber-iii/
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rob(self, root: Optional[TreeNode]) -> int:
@cache
def dfs(root,robbed):
if not root:return 0
if robbed:
return dfs(root.left,False)+dfs(root.right,False)+root.val
return max(dfs(root.left,False),dfs(root.left,True))+max(dfs(root.right,False),dfs(root.right,True))
return max(dfs(root,False),dfs(root,True))
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rob(self, root: Optional[TreeNode]) -> int:
@cache
def dfs(root):
if not root:return [0,0] # 偷,不偷
left = dfs(root.left)
right = dfs(root.right)
steal = max(left[0],left[1])+max(right[0],right[1])
return [left[1]+right[1]+root.val,steal]
res = dfs(root)
return max(res[0],res[1])