python实现单向链表

#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
        """
        nHead = ListNode(0)
        carry = 0
        head = nHead
        flag = 0
        while l1 or l2:
            node = ListNode(flag)
            print(node.val,node.next)
            if l1:
                node.val += l1.val
                l1 = l1.next
            if l2:
                node.val += l2.val
                l2 = l2.next
            flag = node.val // 10
            node.val %= 10
            head.next, head = node, node
            print(node.val,node.next,flag,head.val,head.next)
        print(head.val,nHead.val,nHead.next.val,nHead.next.next.val)
        return nHead.next
if __name__=="__main__":
    nl = ListNode(0)
    nl2 = ListNode(0)
    l0 = nl
    l1 = nl
    list1=[5,8,6,7]
    for x in list1:
        node = ListNode(0)
        node.val = x
        l0.next=node
        l0 = node
    print (nl2.next,nl.next.next.val,nl.next.next.next.val,l1.next.next.next.next.val)

本来就对c中的链表不熟悉，做leetcode第二题直接看蒙，链表可以通过 listcode.val 访问当前值，通过listcode.next 指向下一个值的地址（类型还是listcode）