代碼随想錄48天| 198.打家劫舍 213.打家劫舍II 337.打家劫舍III

198.打家劫舍

題目連結：https://leetcode.cn/problems/house-robber/

這道題目還是比較簡單的，經典五部曲

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

213.打家劫舍II

題目連結：https://leetcode.cn/problems/house-robber-ii/

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0
        elif len(nums) == 1:
            return nums[0]
        
        val1 = self.robb(nums[1:])
        val2 = self.robb(nums[:-1])

        return max(val1, val2)

    def robb(self, nums):
        dp = [0] * len(nums)
        dp[0] = nums[0]

        for i in range(1, len(nums)):
            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]
           

337.打家劫舍III

題目連結：https://leetcode.cn/problems/house-robber-iii/

這道題目剛開始就直接暴力遞歸了，時間過不去，是以放題解

# 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:
        dp = self.traversal(root)
        return max(dp)
    
    def traversal(self, node):
        if not node:
            return (0, 0)

        left = self.traversal(node.left)
        right = self.traversal(node.right)

        # 不偷目前節點, 偷子節點
        val_0 = max(left[0], left[1]) + max(right[0], right[1])

        # 偷目前節點, 不偷子節點
        val_1 = node.val + left[0] + right[0]

        return (val_0, val_1)