题目

给定两个单链表，链表的每个结点代表一位数，计算像个数的和，例如：
3>1>5 和 5>9>2 输出 8>0>8,注意个位数在链表头。

#%%
class LHead:
    def __init__(self):
        self.next = None
    
    def get_ordered_list(self): # 以列表形式打印顺序表
        order_list = []
        if self.next == None or self == None:
            return order_list
        else:
            cur = self.next
            while cur != None:
                order_list.append(cur.data)
                cur = cur.next
            return order_list
        
    def ordered_list_size(self): #获得链表长度
        
        return len(self.get_ordered_list())
    
    def create_ordered_list(self,size): #在新建头结点后面生成有序链表
        i = 0
        tmp = None
        cur = self
        ordered_list = []
         #构造单链表
        while i < size:
             tmp = LNode(i)
             tmp.next = None
             cur.next = tmp
             cur = tmp
             i += 1
        cur = self.next
        while cur != None:
            ordered_list.append(cur.data)
            cur = cur.next
        return ordered_list,self
    
#%%
class LNode:
    def __init__(self,data):
        self.data = data
        self.next = None


#%%
# 整数相加法，缺点是当数字较大时（超出了long的表示范围，就无法使用这种方法）
def head_to_num1(head):
    if head == None or head.next == None:
        return 0
    else:
        list1 = []
        num = 0
        cur = head.next
        while cur != None:
            list1.append(cur.data)
            cur = cur.next
        list1.reverse()
        for i in list1:
            num = str(num) + str(i)
        num = int(num)
    return num


def head_to_num2(head):
    if head == None or head.next == None:
        return 0
    else:
        list1 = []
        num = 0
        cur = head.next
        while cur != None:
            list1.append(cur.data)
            cur = cur.next
        list1.reverse()
        for idx, value in enumerate(list1):
            num = num + value * 10 **(len(list1) - idx - 1)
    return num

def head_num_add(head1,head2):
    num_list = []
    head = LHead()
    cur = head
    num1 = head_to_num2(head1)
    num2 = head_to_num2(head2) # 使用第二种字符串变为数字的函数
    nums = num1 + num2
    nums = str(nums)
    for i in nums:
        num_list.append(int(i))
    num_list.reverse()
    for i in num_list:
        a = LNode(i)
        cur.next = a
        cur = cur.next
    return head
        
#%%
# 链表相加法
def head_add_LHead(head1,head2):
    if head1 == None or head1.next == None:
        return head2
    if head2 == None or head2.next == None:
        return head1
    
    plus = 0
    cur1 = head1.next
    cur2 = head2.next
    sum_head = LHead()
    sum_cur = sum_head 
    while cur1 != None and cur2 != None:
        sums = cur1.data + cur2.data + plus
        if sums > 9:
            plus = 1
            sums = sums - 10
        else:
            plus = 0
        a = LNode(sums)
        sum_cur.next = a
        sum_cur = sum_cur.next
        cur1 = cur1.next
        cur2 = cur2.next
    if cur1:
        cur = cur1
    else:
        cur = cur2
    while cur != None:
        sums = cur.data + plus
        if sums > 9:
            plus = 1
            sums = sums - 10
        else:
            plus = 0
        b = LNode(sums)
        sum_cur.next = b
        sum_cur = sum_cur.next
        cur = cur.next
    if plus == 1:
        sum_cur.next = LNode(1)
    return sum_head
    ```