LeetCode打卡Task07-双指针篇

一、知识点介绍

二、例题

1.反转链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        p1,p2=head,None
        while p1:
            temp=p1.next
            p1.next=p2
            p2=p1
            p1=temp
        return p2

2.删除链表的倒数第 N 个结点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        p1=head
        while n:
            p1=p1.next
            n-=1
        p2=head
        if not p1:
            return p2.next
        while p1.next:
            p2=p2.next
            p1=p1.next
        p2.next=p2.next.next
        return head

3.删除排序链表中的重复元素

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        p1,p2=head,head.next
        while p2:
            flag=0
            while p2 and p1.val==p2.val:
                p2=p2.next
                flag=1
            if flag:
                p1.next=p2
            if not p2:
                break
            p1=p2
            p2=p2.next
        return head

4.环形链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        if not head:
            return False
        p1=p2=head
        while p1.next and p2.next and p2.next.next:
            p1=p1.next
            p2=p2.next.next
            if p1==p2:
                return True
        return False

5.排序链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def findmid(self,head):
        slow=fast=head
        while fast.next and fast.next.next:
            slow=slow.next
            fast=fast.next.next
        return slow
    def merge(self,l,r):
        head=ListNode(None)
        cur=head
        while l and r:
            if l.val<=r.val:
                cur.next=l
                l=l.next
            else:
                cur.next=r
                r=r.next
            cur=cur.next
        cur.next=l or r
        return head.next
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next:
            return head
        mid=self.findmid(head)
        left=head
        right=mid.next
        mid.next=None
        l=self.sortList(left)
        r=self.sortList(right)
        return self.merge(l,r)