给定单向链表，返回翻转后的链表
# coding: utf-8
# @author zhenzong
# @date 2018-05-21 11:07
class Node(object):
def __init__(self, value, next_node):
self.value = value
self.next_node = next_node
def __str__(self):
return str(self.value)
__repr__ = __str__
def revert_list(head):
if not head:
return None
new_head, iter_node = head, head.next_node
head.next_node = None
while iter_node:
tmp = iter_node.next_node
iter_node.next_node = new_head
new_head = iter_node
iter_node = tmp
return new_head
def list_to_str(_node):
if not _node:
return ''
ret = str(_node)
tmp = _node.next_node
while tmp:
ret = '%s,%s' % (ret, tmp)
tmp = tmp.next_node
return ret
def test(length_array):
for length in length_array:
node = None
for i in range(length, 0, -1):
node = Node(i, node)
print 'length: %s, list: %s, revert_list: %s' % (length, list_to_str(node), list_to_str(revert_list(node)))
test([1, 5, 10])
# 输出
# length: 1, list: 1, revert_list: 1
# length: 5, list: 1,2,3,4,5, revert_list: 5,4,3,2,1
# length: 10, list: 1,2,3,4,5,6,7,8,9,10, revert_list: 10,9,8,7,6,5,4,3,2,1
