【LeetCode】两数之和II

1 题目

2 思想

翻转链表后再做计算，然后再翻转链表。本题要求对链表的翻转操作比较熟练。

3 代码

# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        l1 = self.reverseLinkList(l1)
        l2 = self.reverseLinkList(l2)
        head_l1 = l1
        # 执行加法
        flag = 0
        head = None
        while(l1 and l2):
            flag ,l1.val = divmod(l1.val + l2.val + flag,10)            
            head = l1
            l1 = l1.next
            l2 = l2.next
        
        if l2:
            l1 = l2
            head.next = l1
        while(l1):
            flag,l1.val = divmod(l1.val+flag,10)
            head = l1
            l1 = l1.next
        
        # 如果最后还有进位
        if flag :
            head.next = ListNode(flag,None)
        return self.reverseLinkList(head_l1)
        
    # 翻转链表
    def reverseLinkList(self,head):
        # 就地翻转
        nxt = head.next
        pre = None
        while(head and nxt):
            tmp = nxt.next 
            head.next = pre # 修改
            nxt.next = head
            pre = head
            head = nxt
            nxt = tmp
        return head # 返回翻转后的链表