【LeetCode】链表专题

160.相交链表

返回两个链表相交的节点

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if not headA and not headB:
            return None
        p1,p2=headA,headB
        while(p1!=p2):
            p1=p1.next if p1 else headB #短的接上长的，循环到长的也走完的时候，短和长的差便补上了。画图便可以理解
            p2=p2.next if p2 else headA
        return p1

19.删除链表的倒数第N个节点

返回链表的头结点。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
'''快慢指针，之间相隔n个元素'''
class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        fast=head
        slow=head
        
        for i in range(n):
            fast=fast.next
        if not fast:
            return slow.next
        while fast.next:
            fast=fast.next
            slow=slow.next
        slow.next=slow.next.next
        return head

21.合并两个有序链表

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

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        prehead = ListNode(0)
        prev = prehead
        while l1 and l2:
            if l1.val <= l2.val:
                prev.next = l1
                l1 = l1.next
            else:
                prev.next = l2
                l2 = l2.next
            prev=prev.next
        prev.next=l1 if l1 is not None else l2
        return prehead.next