【leetcode】Add Two Numbers

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/MrLJC/p/4240488.html

本文介绍了一种使用链表实现两个非负数相加的方法。通过将数字存储为逆序链表，然后逐位相加并处理进位，最终返回新的链表作为结果。示例输入为 (2->4->3) + (5->6->4)，输出为 7->0->8。

题目描述：

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

解题思路：

链表操作

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class ListNode:
     def __init__(self, x):
         self.val = x
         self.next = None
class Solution:
    # @return a ListNode
    def addTwoNumbers(self, l1, l2):
        l = ListNode(-1)
        lt = l
        jin = 0 
        while l1 != None or l2 != None:
            if l1 == None:
                l1 = ListNode(0)
            if l2 == None:
                l2 = ListNode(0)
            if l.val == -1:
                t = l1.val + l2.val
                if t > 9:
                    jin = 1
                    lt.val = t - 10
                else:
                    lt.val = t
            else:
                t = l1.val + l2.val + jin
                print t
                jin = 0
                if t > 9:
                    jin = 1
                    lt.next = ListNode(t - 10)

                else:
                    lt.next = ListNode(t)
                lt = lt.next
            l1 = l1.next
            l2 = l2.next
        if jin == 1:
            print jin
            lt.next = ListNode(1)
        return l

转载于:https://www.cnblogs.com/MrLJC/p/4240488.html