day2和day3回家太晚，刷完题忘记写笔记了 - -！
24.两两交换链表中的节点
给自己的笔记：
虚拟节点法是创建一个节点，它的next指针指向链表的头节点，这样便于：
- current指向虚拟节点，然后对链表进行操作交换
- 最后返回头节点：return dummyNode.next
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):def swapPairs(self, head):""":type head: ListNode:rtype: ListNode"""#创建虚拟节点dummyNode=ListNode(next=head)#curr 指向dummyNodecurr=dummyNodewhile(curr.next and curr.next.next):#cur的下一位和下下一位都在 才会进入循环temp=curr.nexttemp1=curr.next.next.nextcurr.next=curr.next.nextcurr.next.next=temptemp.next=temp1curr=curr.next.next#curr往前2位return dummyNode.next #返回头节点19 删除链表的倒数第N个节点
- 笨办法，先遍历所有节点获得链表长度。
class Solution(object):def removeNthFromEnd(self, head, n):""":type head: ListNode:type n: int:rtype: ListNode"""#虚拟头节点dummyNode=ListNode(next=head)curr=dummyNodesizeLink=0count=0while(curr.next):sizeLink+=1curr=curr.nextx=sizeLink-ncurr=dummyNodeif(sizeLink==1):return Noneif(sizeLink==n):return head.nextwhile(curr.next):curr=curr.nextcount+=1if(count==x):curr.next=curr.next.next return dummyNode.next
- 双指针法
tips：数组也是可以这样的思路
class Solution(object):def removeNthFromEnd(self, head, n):dummyNode=ListNode(next=head)slow=dummyNodefast=slowcurr=dummyNodefor i in range(n+1):#注意是n+1，这样快指针走到尾，慢指针停在要删的节点前一个节点fast=fast.nextwhile(fast!=None):fast=fast.nextslow=slow.nextslow.next=slow.next.nextreturn dummyNode.next
面试题 02.07. 链表相交
核心思想是求两个链表长度差，让指向长链表头指针移动到，和短链表头指针对齐的位置
比如：
A: [4,1,8,4,5]
currA->4
B:[5,0,1,8,4,5]
currB->0
class Solution(object):def getIntersectionNode(self, headA, headB):""":type head1, head1: ListNode:rtype: ListNode"""dummyNodeA=ListNode(next=headA)currA=dummyNodeAsizeA,sizeB=0,0dummyNodeB=ListNode(next=headB)currB=dummyNodeBcurr,curr2=headA,headBwhile curr:curr=curr.nextsizeA+=1while curr2:curr2=curr2.nextsizeB+=1diff=abs(sizeA-sizeB)#长度差for i in range(diff):#让长链表指针移动到，和短链表头指针对齐的位置if sizeA>sizeB:currA=currA.nextelse:currB=currB.nextwhile currA=currB #当指针相等currA=currA.nextcurrB=currB.nextreturn currA142.环形链表II
…待补充
