单链表反转
唯一就是有一处笔误。
3.4.4 两个链表操作/链表反转 书中代码是：
def rev(self):

p = None
while self._head is not None:
    q = self._head
    self._head = q.next # 摘下原来的原结点
    q._next = p 
    p = q               # 将刚摘下的结点加入 p 引用的结点序列
self._head = p          # 反转后的结点序列已经做好，重置表头链接

其中倒数第三行的 ._next 应该改为 .next，因为结点类 LNode 中类属性是.next。

    q.next = p

如果把其中的 p 变量更名为 tmp_list_head，把 q 改为 sliced_node可能更能理解。
完整的程序，包括测试如下：

coding:utf8

class LinkedListUnderflow(ValueError):

pass

class LNode(object):

def __init__(self, elem, next_=None):
    self.elem = elem
    self.next = next_

class LList(object):

def __init__(self):
    self._head = None

def is_empty(self):
    return self._head is None

def prepend(self, elem):
    self._head = LNode(elem, self._head)

def pop(self):
    if self._head is None:
        raise LinkedListUnderflow("in pop")
    e = self._head.elem
    self._head = self._head.next
    return e

def append(self, elem):
    if self._head is None:
        self._head = LNode(elem)
        return
    p = self._head
    while p.next is not None:
        p = p.next
    p.next = LNode(elem)

def pop_last(self):
    if self._head is None:
        raise LinkedListUnderflow
    p = self._head
    if p.next is None:
        e = p.elem
        self._head = None
        return e
    while p.next.next is not None:
        p = p.next
    e = p.next.elem
    p.next = None
    return e

def rev(self):
    tmp_list_head = None
    while self._head is not None:
        sliced_node = self._head
        self._head = self._head.next
        sliced_node.next = tmp_list_head
        tmp_list_head = sliced_node 
    self._head = tmp_list_head

def elements(self):
    p = self._head
    while p is not None:
        yield p.elem
        p = p.next

def main():

lst = LList()
for i in range(10):
    lst.append(i)
print "original linked list:"
for i in lst.elements():
    print i

print '-' * 25
print "reserved linked list:"

lst.rev()
for i in lst.elements():
    print i 

if name == '__main__':

main()