python与算法：单链表剖分函数(对链表的元素可以按照是否满足特定功能切分为两个新的链表)

def funct(x):
    if x%2==0:
        return True
    else:
        return False
def partition(lst,pred):
    satisfy_list=LList()
    unsatisfy_list=LList()
    p=lst._head
    # lst里面有元素
    while p.next is not None:
        elem=p.elem
        # 如果满足条件，加入到
        if pred(elem):
            if satisfy_list._head is None:
                satisfy_list._head=LNode(elem)
            q=satisfy_list._head
            while q.next is not None:
                q=q.next
            q.next=LNode(elem)
        else:
            if unsatisfy_list._head is None:
                unsatisfy_list._head=LNode(elem)
            r=unsatisfy_list._head
            while r.next is not None:
                r=r.next
            r.next=LNode(elem)
        p=p.next
    satisfy_list.printall()
    unsatisfy_list.printall()
    return satisfy_list,unsatisfy_list
# 测试，程序需要继承前文的LList类
# 创造一个链表加入元素
m1=LList()
for i in range(20):
    m1.append(i)
m2=LList()
for i in range(11,20):
    m2.append(i)
print('m1',m1.printall())
print('m2',m2.printall())
x,y=partition(m2,funct)
print(m2.printall())
print(x.printall())
print(y.printall())