网站如何做宣传百度注册网站怎么弄
198. 打家劫舍
自己的思路：
初始化两个dp数组，dp[i][0]表示不偷第i户，在0-i户可以偷到的最大金额，dp[i][1]表示偷i户在0-i户可以偷到的最大金额。
class Solution:def rob(self, nums: List[int]) -> int:n = len(nums)dp = [[0] * 2 for _ in range(n)]dp[0][1] = nums[0]for i in range(1, n):dp[i][0] = max(dp[i - 1][0], dp[i - 1][1])dp[i][1] = dp[i - 1][0] + nums[i]return max(dp[n - 1][0], dp[n - 1][1])优化：
有一点臃肿，可以优化。dp[i][1]实际上跟dp[i-1][1]就没啥关系，直接把dp数组初始化成一维的就行了。
class Solution:def rob(self, nums: List[int]) -> int:n = len(nums)if n == 1:return nums[0]if n == 2:return max(nums[0], nums[1])dp = [0] * ndp[0] = nums[0]dp[1] = max(nums[0], nums[1])for i in range(2, n):dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])return dp[n - 1]更优化：
想起了斐波那契数列...
class Solution:def rob(self, nums: List[int]) -> int:n = len(nums)if n == 1:return nums[0]if n == 2:return max(nums[0], nums[1])prev = nums[0]cur = max(nums[0], nums[1])for i in range(2, n):prev, cur = cur, max(cur, prev + nums[i])return cur213. 打家劫舍 II
还是比较容易想到的，把环展成两个线性的，一个去头一个去尾即可。
class Solution:def rob(self, nums: List[int]) -> int:n = len(nums)if n == 1:return nums[0]if n == 2:return max(nums[0], nums[1])def helper(n, nums):dp = [0] * ndp[0] = nums[0]dp[1] = max(nums[0], nums[1])for i in range(2, n):dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])return dp[n - 1]return max(helper(n-1, nums[1:]), helper(n-1, nums[:-1]))337. 打家劫舍 III
我的思路：
有点类似贪心的最后一题。
一顿操作AC了，中间遗漏了几种情况，修改后正确了。
class Solution:def rob(self, root: Optional[TreeNode]) -> int:def helper(root):if not root.left and not root.right:return 0, root.valelif root.left and root.right:left_without_self, left_with_self = helper(root.left)right_without_self, right_with_self = helper(root.right)return max(left_without_self + right_without_self, left_with_self + right_with_self, left_without_self + right_with_self, left_with_self + right_without_self), left_without_self + right_without_self + root.valelif root.left and not root.right:left_without_self, left_with_self = helper(root.left)return max(left_with_self, left_without_self), left_without_self + root.valelif root.right and not root.left:right_without_self, right_with_self = helper(root.right)return max(right_with_self, right_without_self), right_without_self + root.valreturn max(helper(root))优化：
终止条件从叶子节点改成空节点，可以将之后的情况全部统一起来。
class Solution:def rob(self, root: Optional[TreeNode]) -> int:def helper(root):if not root:return 0, 0left = helper(root.left)right = helper(root.right)not_include = max(left) + max(right)include = left[0] + right[0] + root.valreturn not_include, includereturn max(helper(root))带备忘的递归：
class Solution:def rob(self, root: Optional[TreeNode]) -> int:memo = {}def helper(node):if not node:return 0if node in memo:return memo[node]val = node.val# 如果偷当前节点，则不能偷其直接的左右子节点，但可以偷其孙子节点if node.left:val += helper(node.left.left) + helper(node.left.right)if node.right:val += helper(node.right.left) + helper(node.right.right)# 不偷当前节点，可以偷其左右子节点not_steal = helper(node.left) + helper(node.right)# 对于当前节点，选择偷与不偷的最大值result = max(val, not_steal)memo[node] = resultreturn resultreturn helper(root)今日总结：
自己写能AC，都能get到要点~~~精简的代码还是得看题解。