视频讲解：动态规划，偷不偷这个房间呢？| LeetCode：198.打家劫舍_哔哩哔哩_bilibili

代码随想录

确定递推公式很重要，在递推公式的提示下可以写出该题
dp[i]表示截至第i个房间偷到的最大金额

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums)==1:
            return nums[0]
        if len(nums)==1:
            return max(nums[0],nums[1])
        dp=[0]*(len(nums))
        dp[0]=nums[0]
        dp[1]=max(nums[0],nums[1])
        for i in range(2,len(nums)):
            dp[i]=max(dp[i-2]+nums[i],dp[i-1])
        return dp[len(nums)-1]

213.打家劫舍II

视频讲解：动态规划，房间连成环了那还偷不偷呢？| LeetCode：213.打家劫舍II_哔哩哔哩_bilibili

代码随想录

在上一题的基础上修改，考虑两种情况，去头或去尾，两种情况返回最大选项

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums)==1:
            return nums[0]
        def rob_sub(nums):
            if len(nums)==0:
                return 0
            if len(nums)==1:
                return nums[0]
            dp=[0]*(len(nums))
            dp[0]=nums[0]
            dp[1]=max(nums[0],nums[1])
            for i in range(2,len(nums)):
                dp[i]=max(dp[i-2]+nums[i],dp[i-1])
            return dp[len(nums)-1]
        case1=rob_sub(nums[:-1])
        case2=rob_sub(nums[1:])
        return max(case1,case2)

337.打家劫舍III

视频讲解：动态规划，房间连成树了，偷不偷呢？| LeetCode：337.打家劫舍3_哔哩哔哩_bilibili

代码随想录

# 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:
        def dfs(node):
            if not node:
                return (0,0)
            left=dfs(node.left)
            right=dfs(node.right)
            not_rob=max(left)+max(right)
            rob=node.val+left[0]+right[0]
            return (not_rob,rob)
        dp=dfs(root)
        return max(dp)