LEETCODE-DAY4

---

title: LEETCODE DAY4
date: 2024-02-25 11:00:40
tags:

DAY2

今日题目：LEETCODE 24（两两交换链表中的节点）、19（删除链表的倒数第N个节点）、160 链表相交 (面试题 02.07.) 、142（环形链表II）
附加题：

T1 两两交换链表中的节点 mark

class ListNode:
     def __init__(self, val=0, next=None):
         self.val = val
         self.next = next
class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy_head=ListNode(next=head)
        cur=dummy_head.next
        pre=dummy_head.next
        while cur.next and cur.next.next:
            cur=cur.next
            temp=cur.next
            pre.next=temp
            cur.next=pre
            pre=temp
            cur=temp
        
        return head

AttributeError: ‘NoneType’ object has no attribute ‘next’
^^^^^^^^
while cur.next and cur.next.next:
Line 16 in swapPairs (Solution.py)

    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy_head=ListNode(next=head)
        cur=dummy_head.next
        pre=dummy_head.next
        while cur and cur.next:
            cur=cur.next
            temp=cur.next
            pre.next=temp
            cur.next=pre
            pre=temp
            cur=temp
        
        return head

输入
head =
[1,2,3,4]
输出
[1,3]
预期结果
[2,1,4,3]

这里逻辑有点不对，相当于是在看1 2 3 的交换（0=dummy_head）所以结果上显示1和2没有交换成功

往后看不行改成往前看（即考虑0 1 2）

class ListNode:
     def __init__(self, val=0, next=None):
         self.val = val
         self.next = next
class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy_head=ListNode(next=head)
        cur=dummy_head.next
        pre=dummy_head
        while cur and cur.next:
            temp=cur.next
            pre.next=temp
            cur.next=temp.next
            temp.next=cur
            pre=cur
            cur=cur.next
           
        return head

输入
head =
[1,2,3,4]
输出
[1,4,3]
预期结果
[2,1,4,3]

修改一下输出
改为return dummy_head.next
成功AC

T2

倒数第n个节点与最后一个节点差（n-1）个脚标

这里fast比slow快n步

fast到最后一个节点时，slow在倒数第n个节点前一节点

class ListNode:
    def __init__(self, val=0, next=None):
         self.val = val
         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        fast=ListNode(next=head)
        slow=ListNode(next=head)
        for i in range(n):
            fast=fast.next
        while fast.next:
            slow=slow.next
            fast=fast.next
        slow.next=slow.next.next
        return head

输入
head =
[1]
n =
1
输出
[1]
预期结果
[]

用例[1,2,3,4,5] n=2，通过

修改

def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy_head=ListNode(next=head)
        fast=slow=dummy_head
        for i in range(n):
            fast=fast.next
        while fast.next:
            slow=slow.next
            fast=fast.next
        slow.next=slow.next.next
        return dummy_head.next

AC

T3

没太搞懂为什么数值相等与指针相等不同（似乎是要看地址）

即：[4,1,8,4,5]和[5,0,1,8,4,5]的相交节点为什么不是1啊？

答：该题交点不是数值相等，而是指针相等！注意是指针哦！

假设指针P1指向[4,1,8,4,5]里的1，地址是001。

假设指针P2指向[5,0,1,8,4,5]里的1，地址是002。

虽然两个指针指向的节点值相同，但是地址不同，所以这并不是同一个节点，也就不是题目的交点。

T4

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        slow=head
        fast=head
        i=0
        while slow :
            fast=slow.next
            while fast:
                if fast !=slow:
                    fast=fast.next
                else:
                    return i
                    break


            slow=slow.next
            i+=1
        

超出时间限制

def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        visited=set()
        while head:
            if head in visited:
                return head
            else:
                visited=visited.add(head)
                head=head.next
        return None

TypeError: argument of type ‘NoneType’ is not iterable
^^^^^^^^^^^^^^^
if head in visited:
Line 11 in detectCycle (Solution.py)

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        visited=set()

        while head:
            if head in visited:
                return head
            visited.add(head)
            head=head.next
        return None

AC