Leetcode 445. 两数相加 II ----python

1. 题目描述

给定两个非空链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储单个数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。
示例:
输入: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出: 7 -> 8 -> 0 -> 7

2. 解题思路

此题若不逆置初始链表，可以用栈来存储链表的值，然后相加，再把相加后的值存在新链表中。
步骤：
（1）用两个栈（stack1，stack2）分别存下两个链表的值
（2）从两个栈中取出元素，相加，相加后放在新的栈（stack3）中
（3）把新栈里的元素放入链表

3.代码实现

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        stack1 = []
        stack2 = []
        stack3 = []
        # 将两链表的元素分别入栈
        while(l1):
            stack1.append(l1.val)
            l1 = l1.next
        while(l2):
            stack2.append(l2.val)
            l2 = l2.next
        lend = 0

        # 将两链表的元素相加后，入栈
        while(stack1 and stack2):
            temp = stack1.pop(len(stack1)-1) + stack2.pop(len(stack2)-1)+ lend
            if(temp < 10):
                stack3.append(temp)
                lend = 0
            else:
                stack3.append(int(str(temp)[1]))
                lend = 1

        while(stack1 == [] and  stack2 != []):
            temp = stack2.pop(len(stack2)-1)+ lend
            if(temp < 10):
                stack3.append(temp)
                lend = 0
            else:
                stack3.append(int(str(temp)[1]))
                lend = 1

        while(stack2 == [] and  stack1 != []):
            temp = stack1.pop(len(stack1)-1) + lend
            if(temp < 10):
                stack3.append(temp)
                lend = 0
            else:
                stack3.append(int(str(temp)[1]))
                lend = 1

        if(lend == 1):
            stack3.append(1)

        #将栈的值放入链表
        dummy = ListNode(None)
        head = dummy
        for i in stack3[::-1]:
            node = ListNode(i)
            head.next = node
            head = head.next
        return dummy.next

4. 测试用例和测试结果

测试用例

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

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        stack1 = []
        stack2 = []
        stack3 = []
        # 将两链表的元素分别入栈
        while(l1):
            stack1.append(l1.val)
            l1 = l1.next
        while(l2):
            stack2.append(l2.val)
            l2 = l2.next
        lend = 0

        # 将两链表的元素相加后，入栈
        while(stack1 and stack2):
            temp = stack1.pop(len(stack1)-1) + stack2.pop(len(stack2)-1)+ lend
            if(temp < 10):
                stack3.append(temp)
                lend = 0
            else:
                stack3.append(int(str(temp)[1]))
                lend = 1

        while(stack1 == [] and  stack2 != []):
            temp = stack2.pop(len(stack2)-1)+ lend
            if(temp < 10):
                stack3.append(temp)
                lend = 0
            else:
                stack3.append(int(str(temp)[1]))
                lend = 1

        while(stack2 == [] and  stack1 != []):
            temp = stack1.pop(len(stack1)-1) + lend
            if(temp < 10):
                stack3.append(temp)
                lend = 0
            else:
                stack3.append(int(str(temp)[1]))
                lend = 1

        if(lend == 1):
            stack3.append(1)

        #将栈的值放入链表（新开辟的链表空间）
        dummy = ListNode(None)
        head = dummy
        for i in stack3[::-1]:
            node = ListNode(i)
            head.next = node
            head = head.next
        return dummy.next

    #打印链表操作
    def printLink(self,root):
        r = root
        while(r):
            print(r.val)
            r = r.next

root1 = ListNode(9)
n2 = ListNode(4)
n3 = ListNode(6)
root1.next = n2
n2.next = n3

root2 = ListNode(0)
# r2 = ListNode(5)
# r3 = ListNode(7)
# root2.next = r2
# r2.next = r3

s = Solution()
result= s.addTwoNumbers(root1,root2)
s.printLink(result)

测试结果：

9
4
6