打家劫舍I
问题描述
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:ec9b2891-2dab-4fc5-8a11-3f6773c8b3d6
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:76bc0246-90eb-4e3b-b644-780b17a03eed
示例:
输入:[1,2,3,1]
输出:4
解释:偷窃 1 号房屋 (金额 = 1) ,然后偷窃 3 号房屋 (金额 = 3)。偷窃到的最高金额 = 1 + 3 = 4 。
分析问题
首先,我们先将问题简化处理。假设目前只有一间房屋,则偷窃该房屋,此时就是偷窃到的最高总金额。如果只有两间房屋,因为此时两间房屋相邻,只能偷窃其中的一间房屋,可以选择其中金额较高的房屋进行偷窃,就是可以偷窃到的最高总金额。如果房屋的数量大于两间,假设小偷此时处于第k(k>2)间房屋,那么他有两个选择。
- 如果他偷窃第k间房屋,那么他就不能偷窃第k-1间房屋,此时其能偷窃到的总金额为前k-2间房屋的最高总金额和第k间房屋的金额之和。
- 如果他不偷窃第k间房屋,那么此时其能偷窃到的总金额为前k-1间房屋的最高总金额。
在上述两个选项中选择金额较大的即为前k间房屋能偷窃到的最高总金额。
我们用 dp[i] 来表示前 i 间房屋能偷窃到的最高总金额,经过前面的分析,可以知道其状态转移方程为:
dp[i] = max( dp[i-2] + nums[i] , dp [i-1])
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:d4252c7a-196c-488b-99e7-4d62fb12725e
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:4f60e491-c08b-454f-8c50-3d1ed44b032e
- 当只有一间房屋时,此时dp[0] = nums[0],表示偷窃该房屋。
- 当只有两间房屋时,此时 dp[1] = max(nums[0] , nums[1]),即在这两间房屋中选择金额较大的房屋进行偷窃。
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:cb4f7370-7318-4ad5-8652-3d461e83ca22
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:8b698041-bb6f-4774-985a-d8f7ce875511
class Solution:
def rob(self, nums):
#如果数组为空,则直接返回0
if not nums:
return 0
length = len(nums)
#如果房屋数量等于1
#则直接偷窃第一间房屋,
#所以此时能偷窃到的最大金额是nums[0]
if length == 1:
return nums[0]
dp = [0] * length
#边界条件
dp[0] = nums[0]
dp[1] = max(nums[0], nums[1])
for i in range(2, length):
dp[i] = max(dp[i - 2] + nums[i], dp[i - 1])
return dp[length - 1]
该算法的时间复杂度是O(n),空间复杂度也是O(n)。
通过观察,我们发现dp[i] 只和 dp[i-2] 和 dp[i-1]有关,即只和该房屋的前两间房屋的最高总金额相关,所以我们可以使用滚动数组,在每个时刻只需要存储前两间房屋的最高总金额即可,从而降低空间复杂度。我们来看一下代码的实现。
class Solution:
def rob(self, nums):
#如果数组为空,则直接返回0
if not nums:
return 0
length = len(nums)
#如果房屋数量等于1
#则直接偷窃第一间房屋,
#所以此时能偷窃到的最大金额是nums[0]
if length == 1:
return nums[0]
#边界条件
first, second = nums[0], max(nums[0], nums[1])
for i in range(2, length):
first, second = second, max(first + nums[i], second)
return second
该算法的时间复杂度是O(n),空间复杂度是O(1)。
打家劫舍II
问题描述
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:b3f9e15d-076d-4f1e-bf75-87fbf84b4aba
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:92c218c2-18e1-4478-948b-da59cf14c399
示例:
输入:nums = [2,3,2]
输出:3
解释:你不能先偷窃 1 号房屋(金额 = 2),然后偷窃 3 号房屋(金额 = 2), 因为他们是相邻的。
分析问题
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:82c0efb5-3a28-45c9-8a59-4ef2dfcac042
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:2b81df58-354e-4f7b-a6cc-d7e5558fe42a
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:e7810c64-a109-4019-a20d-4658c8822309
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:e8119b20-7613-4133-ade0-aa729072f068
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:0f24136d-a42c-452f-b438-e7fbe2ce5c7b
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:7c800bac-89b6-4bfe-b35e-1e6f248b0561
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:93faece7-59ad-4cf4-a184-98498586d0cc
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:4bb750f0-f069-4e23-911a-d17aa9fa4b50
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:edfc0ecd-c8c3-473e-872b-b212c4a91218
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:20ce6f93-08ff-4651-9c9e-1b1c19061d28
我们假设数组 nums 的长度为n。如果不偷窃最后一间房屋,则可以偷窃的房屋的下标是0n-2;如果不偷窃第一间房屋,则可以偷窃的房屋的下标是1n-1。
接下来我们就可以采用上一题的解法,对于两段下标范围分别计算可以偷窃到的最高总金额,其中的最大值即为在 n 间房屋中可以偷窃到的最高总金额。
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:733e1838-ebe8-4a45-a095-a963df6b86b7
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:dd782f4c-dec8-4505-b15c-dc940c8c2ddb
class Solution:
def rob(self, nums):
#求nums[start,end]范围内可以偷窃到的最大金额
def help(start, end):
first = nums[start]
second = max(nums[start], nums[start + 1])
for i in range(start + 2, end + 1):
first, second = second, max(first + nums[i], second)
return second
length = len(nums)
#边界条件
if length == 1:
return nums[0]
elif length == 2:
return max(nums[0], nums[1])
else:
return max(help(0, length - 2), help(1, length - 1))
该算法的时间复杂度是O(n),空间复杂度是O(1)。
打家劫舍III
问题描述
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:02343fcb-65f9-45e5-9a1d-1b0de629ad8f
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:b3d78edb-82b2-44fd-9a6a-734967df665d
示例:
输入:[3,2,3,null,3,null,1]
3
/ \
2 3
\ \
3 1
输出:7
解释:小偷一晚能够盗取的最高金额 = 3 + 3 + 1 = 7。
分析问题
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:ed41d2f9-a42e-493d-8475-82117c89e5fe
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:4dbe6535-1e5b-4fbd-b17e-e055e0677223
首先我们用f(a)来表示选择a节点的情况下,a节点的子树上被选择的节点的最大权值和。g(a)表示在不选择a节点的情况下,a节点的子树上被选择的节点的最大权值和。l 和 r 分别表示a的左右孩子。
小偷对于树中的每个节点都有偷或者不偷两种选择,假设当前的节点是a。
- 当a被选中时,a的左右孩子都不能被选中,所以a被选中的情况下,它的子树上被选择的节点的最大权值和为l 和 r 不被选中的最大权值和相加,即 f(a) = g(l) + g(r)。
- 当a不被选中时,a的左右孩子可以被选中,也可以不被选中。此时 g(a) = max { f(l) , g(l) } + max{ f(r) , g(r) }。
这里我们可以使用深度优先搜索的办法后序遍历这棵二叉树,就可以得到每一个节点的 f 和 g。根节点的 f和 g 的最大值就是我们要找的答案。
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:db553247-8671-40cc-8e18-fa5710b4c1d8
[En]
[TencentCloudSDKException] code:FailedOperation.ServiceIsolate message:service is stopped due to arrears, please recharge your account in Tencent Cloud requestId:5f7228a0-5bc4-48e2-9d51-d19068fd18c6
class Solution:
def __init__(self):
self.f={}
self.g={}
def dfs(self,node):
if not node:
return
self.dfs(node.left)
self.dfs(node.right)
#表示选中该节点
self.f[node]=node.val + self.g.get(node.left,0) + self.g.get(node.right,0)
#表示没有选择该节点
self.g[node] = max(self.f.get(node.left,0),self.g.get(node.left,0)) \
+ max(self.f.get(node.right,0),self.g.get(node.right,0))
def rob(self, root):
self.dfs(root)
return max(self.f.get(root,0),self.g.get(root,0))
该算法的时间复杂度是O(n),空间复杂度也是O(n)。
这里我们还可以优化一下,因为无论是求 f(a) 还是 g(a),他们的值只和 f(l) 、g(l)、f(r)和g(r)有关。所以对于每一个节点,我们只关心它的左右孩子的 f 和 g 是多少。在python中,我们可以用元组来表示,每次递归返回的时候,都把这个点对应的 f 和 g 返回给上一级调用。这样我们就可以省去哈希表的空间,下面我们来看一下具体的代码实现。
class Solution:
def dfs(self,node):
if not node:
return (0,0)
left=self.dfs(node.left)
right=self.dfs(node.right)
#表示选中该节点
selected = node.val + left[1] + right[1]
#表示没有选择该节点
noselected = max(left[0],left[1]) \
+ max(right[0],right[1])
return (selected,noselected)
def rob(self, root):
result=self.dfs(root)
return max(result[0],result[1])
该算法的时间复杂度是O(n),空间复杂度也是O(n)。
Original: https://www.cnblogs.com/cxyxz/p/15599580.html
Author: 算法推荐管
Title: 面试常遇的打家劫舍问题你学会了吗~
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