难度：中等

题目内容：

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶：

如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例：

输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7

题解:

emmm
首先最简单的情况就是有数为0 ，直接返回另一个就好了
其他的情况可以先把列表翻转过来，然后相加的出一个新列表，把新列表再翻转一下就得到结果了。
不过题目要求不能进行翻转
emmm那也好说，我们加一个位数计数就好了
不过要注意的是（反正我出错了）相加是可能进多位的（199+1这样）
我下面这个就算初步修改了一下，但是只能解决进两位。。。所以打算换种方法，先算出来，然后把中间变成两位的数拆开就好了嘛，比如10拆成1 0

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        ###如果有0的话
        if l1.val == 0:
            return l2
        elif l2.val == 0:
            return l1
        l1len = 0
        l1c = l1
        while l1c:
            l1c = l1c.next
            l1len += 1
        l2len = 0
        l2c = l2
        while l2c :
            l2c = l2c.next
            l2len += 1
        if l2len >= l1len:
            return self.sum(l2,l2len,l1,l1len)
        elif l2len < l1len:
            return self.sum(l1,l1len,l2,l2len)

    def sum(self,lbig,lbiglen,lsmall,lsmalllen):
        # if lbiglen == lsmalllen and (lbig.val +lsmall.val) >= 10:
        #     r = ListNode(1)
        r0 = ListNode(0)
        r = r0
        while lbig:
            if lbiglen == lsmalllen:
                if  (lbig.val +lsmall.val) >= 10:
                    r.val += 1

                r.next = ListNode(mod( (lbig.val +lsmall.val) ,10))
                if r.next.val == 9:#预警一下会不会连进两位
                    temp = r
                r = r.next
                lbig = lbig.next
                lbiglen -= 1
                lsmall = lsmall.next
                lsmalllen -= 1
            else:
                r.next = ListNode(lbig.val)
                if r.next.val == 9:#预警一下会不会连进两位
                    temp = r
                r = r.next
                lbig = lbig.next
                lbiglen -= 1
        if r0.val == 0 and r0.next:
            r0 = r0.next
        return r0

稍加改动，就可以通过啦！

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        ###如果有0的话
        if l1.val == 0:
            return l2
        elif l2.val == 0:
            return l1
        l1len = 0
        l1c = l1
        while l1c:
            l1c = l1c.next
            l1len += 1
        l2len = 0
        l2c = l2
        while l2c :
            l2c = l2c.next
            l2len += 1
        if l2len >= l1len:
            return self.sum(l2,l2len,l1,l1len)
        elif l2len < l1len:
            return self.sum(l1,l1len,l2,l2len)

    def sum(self,lbig,lbiglen,lsmall,lsmalllen):
        # if lbiglen == lsmalllen and (lbig.val +lsmall.val) >= 10:
        #     r = ListNode(1)
        r0 = ListNode(0)
        r = r0
        while lbig:
            if lbiglen == lsmalllen:
                if  (lbig.val +lsmall.val) >= 10:
                    r.val += 1
                r.next = ListNode(mod( (lbig.val +lsmall.val) ,10))
                r = r.next
                lbig = lbig.next
                lbiglen -= 1
                lsmall = lsmall.next
                lsmalllen -= 1
            else:
                r.next = ListNode(lbig.val)
                r = r.next
                lbig = lbig.next
                lbiglen -= 1
        if r0.val == 0 and r0.next:
            r0 = r0.next
        # 拆数
        
        l = lbiglen

        r = r0
        carry = True
        while carry:
            carry = False
            r = r0
            while r.next:
                if r.next.val >= 10:
                    r.val += 1
                    r.next.val -= 10
                    carry = True
                r = r.next
        # 第一位大于等于10要变成两个数
        if r0.val >= 10:
            newNode = ListNode(mod(r0.val,10))
            newNode.next = r0.next
            r0.val = r0.val // 10
            r0.next = newNode
            l += 1
        return r0