擅长:python、mysql、java
<p>@本尼·麦克尼
我找到了另一种方法,使用setattr:</p>
<pre><code>class Node():
def __init__(self, value = None, nex = None):
self.val = value
self.nex = nex
class List(Node):
def __init__(self):
self.head = Node()
def ins(self):
li = []
for i in range(10):
setattr(self,'n' + str(i), Node(i))
#self.n0.nex = self.n1
#self.n1.nex = self.n2
#self.n2.nex = self.n3
#.
#.
#.
#self.n8.nex = self.n9
</code></pre>
<p>现在我有9个节点从self.n0到self.n9
如何将它们连接起来以获得:
self.head.nex=自身n0
self.n0.nex=self.n1
等。。。在</p>