标签：还需 ini nbsp next ack word on() cannot contain
You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7
给定两个整数，求和。注意链表每个结点只放一个数字。高位在前。
思路：两个链表分别进栈，然后出栈时相加，注意设置一个临时变量用来存放进位。每次相加时应该是3个值相加：链表a+链表b+进位。
此外注意的是若最高为还有进位，则继续添加新节点。
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: 9 stack1 = [] 10 stack2 = [] 11 12 newhead = ListNode(0) 13 while(l1): 14 stack1.append(l1.val) 15 l1 = l1.next 16 while(l2): 17 stack2.append(l2.val) 18 l2 = l2.next 19 p = 0 # 进位 20 while stack1 or stack2 or p: # 注意这里别丢了 or p命令。如果有进位则还需创建新节点 21 temp = (stack1.pop() if stack1 else 0) + 22 (stack2.pop() if stack2 else 0) + p # 每次的和为两个链表的和+进位 23 p = temp//10 # 更新进位 24 25 node = ListNode(temp%10) 26 node.next = newhead.next 27 newhead.next = node 28 29 return newhead.next
验证：
1 # 新建链表1 2 listnode1 = ListNode_handle(None) 3 s1 = [1,2,3,4,5,6,7,8] 4 for i in s1: 5 listnode1.add(i) 6 listnode1.print_node(listnode1.head) 7 8 # 新建链表2 9 listnode2 = ListNode_handle(None) 10 s2 = [1,5,9,9,9] 11 for i in s2: 12 listnode2.add(i) 13 listnode2.print_node(listnode2.head) 14 15 s = Solution() 16 head = s.addTwoNumbers(listnode1.head, listnode2.head) 17 listnode1.print_node(head)
1 2 3 4 5 6 7 8
1 5 9 9 9
1 2 3 6 1 6 7 7
标签：还需 ini nbsp next ack word on() cannot contain
原文地址：https://www.cnblogs.com/king-lps/p/10664503.html
