编写一个程序，删除单链表中数据域重复的节点

这里给出一个Python的实现：

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def delete_duplicate(head):
    if head is None:
        return
    current = head
    while current is not None:
        runner = current
        while runner.next is not None:
            if runner.next.data == current.data:
                runner.next = runner.next.next
            else:
                runner = runner.next
        current = current.next

# 测试
head = Node(1)
n2 = Node(2)
n3 = Node(3)
n4 = Node(2)
n5 = Node(4)
head.next = n2
n2.next = n3
n3.next = n4
n4.next = n5

delete_duplicate(head)

current = head
while current is not None:
    print(current.data)
    current = current.next

输出结果为：

1
2
3
4

思路解析：

这个问题可以用两个指针current和runner来解决。current指向当前节点，runner从current开始向后遍历，如果发现runner指向的节点的数据和current指向的节点的数据相同，就将runner指向的节点删除；否则，runner继续向后遍历。当runner遍历完链表后，current才向后移动一位，继续进行上述操作，直到current遍历完链表为止。这样就能够删除单链表中数据域重复的节点了。