Problem 2: Add Two Numbers

最新推荐文章于 2025-09-06 22:26:52 发布

AlexYan91

最新推荐文章于 2025-09-06 22:26:52 发布

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分类专栏： LeetCode 文章标签： leetcode python 指针

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本文介绍了使用Python实现两个链表表示的非负整数相加的算法过程。通过将链表中的数字以逆序存储，并逐位相加的方式，最终返回新的链表结果。文章分享了从学习Python到完成算法的全过程。

第二题：

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

Hide Tags Linked List Math

花了一个小时学了下python~感觉python好爽。。。。

第一个算法写得磕磕绊绊。。。代码乱得一塌糊涂：

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

    def __str__(self):
        temp = self
        ret = None
        while temp != None:
            if ret == None:
                ret = str(temp.val)
            else:
                ret = ret+"->"+str(temp.val)
            temp = temp.next
        return ret

class Solution:
    # @return a ListNode
    def addTwoNumbers(self, l1, l2):
        t1 = l1
        t2 = l2

        ret = None
        plusone = False
        while t1 != None or t2 != None:
            p = ListNode(0)
            if ret == None:
                ret = p
                pre = p
            else:
                pre.next = p

            if t1 != None:
                p.val = p.val + t1.val
            if t2 != None:
                p.val = p.val + t2.val

            if plusone:
                p.val = p.val + 1
                plusone = False

            if p.val > 9:
                p.val = p.val - 10
                plusone = True

            pre = p
            if t1 != None:
                t1=t1.next
            if t2 != None:
                t2=t2.next
        else:
            if plusone:
                p = ListNode(1)
                pre.next = p

        return ret

后面改造了一下：

class Solution:
    # @return a ListNode
    def addTwoNumbers(self, l1, l2):

        ret = None
        p3 = None
        p1 = l1
        p2 = l2

        carry = False

        while p1 != None or p2 != None:
            count = 0
            if p1 != None:
                count += p1.val
                p1 = p1.next
            if p2 != None:
                count += p2.val
                p2 = p2.next

            if carry:
                count += 1
                carry = False
            if count > 9:
                count -= 10
                carry = True

            if p3 == None:
                p3 = ListNode(count)
                ret = p3
            else:
                p3.next = ListNode(count)
                p3 = p3.next

        else:
            if carry:
                p3.next = ListNode(1)


        return ret

回顾一下指针的感觉~