合并两个有序链表
解题思想：
分别用l1,l2遍历两个链表，如果l1.data小于l2.data，则将l1归入合并后的链表，l1=l1.next;反之将l2归入合并后的链表，l2=l2.next;如果其中一个链表遍历结束，则将没遍历完的另一链表的剩下部分归入合并后的链表
代码实现：
# -*-coding:utf-8-*-
"""
@Author : 图南
@Software: PyCharm
@Time : 2019/9/7 14:23
"""
class Node:
def __init__(self, data=None, next=None):
self.data = data
self.next = next
def print_link(head):
cur = head.next
while cur.next != None:
print(cur.data, end=' ')
cur = cur.next
print(cur.data)
def con_link(n):
nums = list(map(int, n.split(' ')))
head = Node()
cur = head
for num in nums:
node = Node(num)
cur.next = node
cur = node
return head
def mergeLink(head1, head2):
head = Node
cur = head
l1 = head1.next
l2 = head2.next
while l1 is not None and l2 is not None:
if l1.data < l2.data:
cur.next = l1
l1 = l1.next
cur = cur.next
cur.next = None
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = None
if l1 is None:
cur.next = l2
else:
cur.next = l1
print_link(head)
if __name__ == '__main__':
n1 = input("Link1:")
n2 = input("Link2:")
head1 = con_link(n1)
head2 = con_link(n2)
mergeLink(head1, head2)
运行结果：
