House Robber I/II/III

这题代表了单向线性dp的基本pattern: build local best value at each element and track the global optimal. 这题的要点是local是前k个房间的最优解，这个解不一定要选第k个房间。这个local解是比较k-1的optimal和k-2 optimal + current value。之所以不以必选第k个房间作local是因为有可能最优解有连续2或多个房间不选，比如100，1，1，100

III:

• binary tree和array的思路是一样的，对于当前root，需要前面两层的结果：left/right subtree的max和left/right的子树的max
• 同时，和array一样，返回的值不是必然包括root的
• 在计算left/right子树的max时，只需要sum，因为root对于两层后只关心sum

class Solution(object):def rob(self, nums):""":type nums: List[int]:rtype: int"""if not nums: return 0pre_2 = nums[0]if len(nums)<2: return pre_2pre_1 = max(nums[0],nums[1])cur = pre_1for i in range(2, len(nums)):cur = max(pre_1, pre_2+nums[i])pre_2=pre_1pre_1=curreturn cur

class Solution(object):def rob(self, nums):""":type nums: List[int]:rtype: int"""def robOneDir(nums):pre_1 = nums[0]pre_2 = max(nums[0], nums[1])for i in range(2, len(nums)):cur = max(pre_2, pre_1+nums[i])pre_1=pre_2pre_2=curreturn pre_2if not nums: return 0if len(nums)<2: return nums[0]if len(nums)==2: return max(nums[0], nums[1])nums1,nums2=nums[0:-1],nums[1:]return max(robOneDir(nums1), robOneDir(nums2))

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution(object):def rob(self, root):""":type root: TreeNode:rtype: int"""def robRec(root):pre_left_1,pre_left_2,pre_right_1,pre_right_2=0,0,0,0if root.left:pre_left_1, pre_left_2 = robRec(root.left)if root.right:pre_right_1, pre_right_2 = robRec(root.right)cur = max(pre_left_2+root.val+pre_right_2, pre_left_1+pre_right_1)return cur, pre_left_1+pre_right_1if not root: return 0maxValue, noUse = robRec(root)return maxValue