剑指offer 链表(下)

目录

• T22-链表中倒数第k个节点
• T24-反转链表
  • 非递归写法 py2
  • 递归写法 py2
• [T35-复杂链表的复制]
• [删除链表中重复的节点]

T22-链表中倒数第k个节点

注意走(k-1)步的循环中，判断p==None就返回，即链表长度小于k的情况

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
 
class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if k<1 or not head:
            return None
        p,p1 = head,head
        for i in range(k-1):
            p = p.next
            if not p:
                return None
        while p.next:
            p = p.next
            p1 = p1.next
        return p1

T24-反转链表

非递归写法 py2

对于看似复杂的问题，看分解的每一部分怎么操作也许是最好的方法。 此题也是，只关注链表的每个节点，这个单向链表的节点，它的后节点指向变为哪里，而不关心多个节点是怎样连接的。

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        # write code here
        if not pHead or not pHead.next:
            return pHead
        ppre,pnext=None,None
        p = pHead
        while p:
            pnext = p.next
            p.next = ppre
            ppre = p
            p = pnext
        return ppre

递归写法 py2

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        # write code here
        return self._reverse(pHead,None)
    def _reverse(self,node,prev=None):
        if not node:
            return prev
        n = node.next
        node.next = prev
        return self._reverse(n,node)

[T35-复杂链表的复制]

[删除链表中重复的节点]