leetcodeSolutionInPython/leetcode2 at master · wwx13/leetcodeSolutionInPython · GitHub

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# Definition for singly-linked list.
# Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
# Output: 7 -> 0 -> 8
# Explanation: 342 + 465 = 807.
class ListNode(object):  #定义点
    def __init__(self, x):
        self.val = x
        self.next = None

class NList(): #定义链表存储结构
    def __init__(self,node):
        self.head=node
    def add_node(self,addedNode):
        tempnode=self.head
        while tempnode.next!=None:
            tempnode=tempnode.next
        tempnode.next=addedNode
        return self
    def print_order_list(self):
        tempnode=self.head
        print(tempnode.val)
        while tempnode.next!=None:
            tempnode = tempnode.next
            print(tempnode.val)


    def get_len(self):
        tempnode=self.head
        length=1
        while tempnode.next!=None:
            tempnode=tempnode.next
            length+=1
        return length
Head1=ListNode(2)
H1=NList(Head1)
H1.add_node(ListNode(4)).add_node(ListNode(3))
# H1.print_order_list()

#342+65=407  这里使用 65因为 它更具有一般性不要全部位数相同的相加
H2=NList(ListNode(5))
H2.add_node(ListNode(6)).add_node(ListNode(4)).add_node(ListNode(3))
# H2.print_order_list()
# print(H2.get_len())
def sum_Nlist(H1,H2):
    result_head=NList(ListNode(-1))
    res_H=result_head.head
    len1=H1.get_len()
    len2=H2.get_len()
    tempnode1=H1.head
    tempnode2=H2.head
    signal_add=False
    while tempnode1!=None and tempnode2!=None:#边界条件为任何一个没有子节点了
        add_res=tempnode1.val+tempnode2.val

        if signal_add==True:
            add_res=add_res+1
        if add_res>=10:
           tempvalue=add_res%10
           signal_add=True
        else:
            tempvalue=add_res
            signal_add=False
        if res_H.val==-1:  #结果的头节点未赋值
            res_H.val=tempvalue
        else:
            result_head.add_node(ListNode(tempvalue))

        tempnode2=tempnode2.next
        tempnode1=tempnode1.next

    if len1>len2:#是第二个短
        longerList=tempnode1
    elif len1==len2 and signal_add==True:
        result_head.add_node(ListNode(1))
        return result_head
    elif len2>len1:
        longerList=tempnode2
    while longerList!=None:

        if signal_add==True:
            tempvalue=longerList.val+1
        else:
            tempvalue=longerList.val
        if tempvalue>=10:
            result_head.add_node(ListNode(tempvalue%10))
            signal_add=True
        else:
            result_head.add_node(ListNode(tempvalue))
            signal_add=False
        longerList=longerList.next
    if signal_add==True:
        result_head.add_node(ListNode(1))
    return  result_head

sum_Nlist(H1,H2).print_order_list()


####上面的方法使用了链表结构，进位通过if else 信号量判断  代码量多 下面仅仅使用ListNode结构，使用add_num（0，1）进位
# Definition for singly-linked list.
# Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
# Output: 7 -> 0 -> 8
# Explanation: 342 + 465 = 807.
class ListNode(object):  #定义点
    def __init__(self, x):
        self.val = x
        self.next = None

head1=ListNode(2)
point1=head1
point1.next=ListNode(4)
point1=point1.next
point1.next=ListNode(5)

head2=ListNode(5)
point2=head2
point2.next=ListNode(6)
def sum_list(head1,head2):
    add_num=0  #要么加1要么加0 进位
    result_head=ListNode(0)
    point=result_head
    tempnode1=head1
    tempnode2=head2
    while tempnode1!=None and tempnode2!=None:
        value=tempnode1.val+tempnode2.val+add_num
        if value>=10:
            value=value%10
            add_num=1
        else:
            add_num=0
        point.next=ListNode(value)
        point=point.next
        tempnode1=tempnode1.next
        tempnode2=tempnode2.next
    if tempnode1!=None:
        tempnode=tempnode1
    elif tempnode1==None and tempnode2==None:
        return result_head
    else:
        tempnode=tempnode2
    while tempnode!=None:
        value=tempnode.val+add_num
        if value>=10:
            value=value%10
            add_num=1
        else:
            add_num=0
        point.next=ListNode(value)
        point=point.next
        tempnode=tempnode.next
    return result_head
print(sum_list(head1,head2).next.next.next.val)