leetcode 两数相加

题目
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给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807
  1. 这道题比较简单，主要是考链表的，下面是自己写的
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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        l3 = None
        pre1 = l1
        pre2 = l2
        pre = l3
        jin = 0									# 进位
        while pre1 and pre2:
            s = pre1.val + pre2.val + jin
            if s >= 10:
                jin = int(s / 10)
                s = s % 10
            else:
                jin = 0
            cur = ListNode(s)
            if pre:
                pre.next = cur
                pre = pre.next
            else:
                l3 = cur
                pre = cur
            pre1 = pre1.next
            pre2 = pre2.next
            
        left = pre1 if pre1 else pre2
        while left:
            s = left.val + jin
            if s >= 10:
                jin = int(s / 10)
                s = s % 10
            else:
                jin = 0
            cur = ListNode(s)
            pre.next = cur
            pre = pre.next
            left = left.next
        if jin != 0:
            cur = ListNode(jin)
            pre.next = cur
        return l3
  1. 下面是参考别人写的，简洁
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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        ret = ListNode(0)
        cur = ret
        carry = 0
        while l1 or l2:
            x = 0 if l1 is None else l1.val
            y = 0 if l2 is None else l2.val
            s = x + y + carry
            carry = int(s / 10)
            cur.next = ListNode(s % 10)
            cur = cur.next
            if l1:
                l1 = l1.next
            if l2:
                l2 = l2.next
        if carry != 0:
            cur.next = ListNode(carry)
        return ret.next