链表相关

def ReverseList(pHead):
        # write code here
         
 
        # write code here
        if not pHead or not pHead.next:
            return pHead
          
        last = None
          
        while pHead:
            tmp = pHead.next
            pHead.next = last
            last = pHead
            pHead = tmp
        return last

题目描述

在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针。 例如，链表1->2->3->3->4->4->5 处理后为 1->2->5

def deleteDuplication(head):
        if head is None:
            return None
        node=ListNode(0)
        node.next=head
        p=node
        cur=head
        while cur:
            if cur.next and cur.next.val==cur.val:
                tmp=cur.next
                while tmp and tmp.val==cur.val:
                    tmp=tmp.next
                p.next=tmp
                cur=tmp
            else:
                p=cur
                cur=cur.next
        return node.next

 题目描述：链表相邻节点进行翻转，不能直接拷贝值

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

def swapPairs(self, head):
        if not head or not head.next:
            return head
        p1=newHead=ListNode(-1)
        newHead.next=head
        
        
        while p1 and p1.next and p1.next.next:
            p0,p1,p2=p1,p1.next,p1.next.next
            p0.next,p1.next,p2.next=p2,p2.next,p1
            
        return newHead.next