T22-鏈表中倒數第k個節點

注意走(k-1)步的循環中，判斷p==None就返回，即鏈表長度小于k的情況

# -*- 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-反轉鏈表

非遞歸寫法 py2

對于看似復雜的問題，看分解的每一部分怎么操作也許是最好的方法。 此題也是，只關注鏈表的每個節點，這個單向鏈表的節點，它的后節點指向變為哪里，而不關心多個節點是怎樣連接的。

# -*- 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

遞歸寫法 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-復雜鏈表的復制]

[刪除鏈表中重復的節點]