单链表及反转（python，go实现）

python实现：

class ListNode(object):
    def __init__(self, val=0, p=None):
        self.value = val
        self.next = p

    def setNext(self, node):
        self.next = node


class LinkedList(object):
    def __init__(self):
        self.head = ListNode()
        self.len = 0

    def insertAfter(self, p: ListNode, val):
        cur = self.head
        newNode = ListNode(val)
        while (1):
            cur = cur.next
            if cur == p:
                newNode.setNext(cur.next)
                cur.next = newNode
                break
            if cur.next == None:
                return False
        return True

    def inertToTail(self, val):
        cur = self.head
        newNode = ListNode(val)
        while (1):
            if cur.next == None:
                cur.setNext(newNode)
                return True
            cur = cur.next

    # 反转
    def reserve(self):
        pre = self.head.next
        cur = self.head.next.next
        pre.next = None

        while 1:
            if cur.next == None:
                self.head.next = cur
                cur.next = pre
                break
            else:
                tmp = cur.next #保留状态
                cur.next = pre
                pre = cur
                cur = tmp

    def __str__(self):
        cur = self.head
        s = "%s->" % str(self.head.value)
        while (1):
            if cur.next == None:
                return s
            else:
                s += "%s->" % str(cur.next.value)
            cur = cur.next


if __name__ == '__main__':
    ll = LinkedList()
    for i in range(1, 20):
        ll.inertToTail(i)
    print(ll)

    ll.reserve()
    print(ll)

反转需注意：反转的时候需要提前记录cur（标记当前位置）的位置，防止cur的next改变导致死循环

go实现：