LeetCode 2. Add Two Numbers

Add Two Numbers

题目

https://leetcode.com/problems/add-two-numbers/

You are given two non-empty linked lists representing two non-negative integers. 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.You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

知识点

1. Python中的链表和使用：

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

l1_1 = ListNode(1)
l1_2 = ListNode(3)
l1_1.next = l1_2
print l1_1.next.val # 3

l1.next == None时为最后一个节点

2. 输出整个链表 , print的原型是print(value, ..., sep=' ', end='\n', file=sys.stdout) 因而可以通过调整end来更改输出格式

from __future__ import print_function

def print_list(l):
    print(l.val, end=" ")
    while l.next is not None:
        print("->", end=" ")
        print(l.next.val, end=" ")
        l = l.next

解题

1. 直接做模拟加法，因为是倒序排列，输出也是倒序，所以一位一位的相加即可

from __future__ import print_function
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        temp_val = l1.val + l2.val
        l3 = ListNode(temp_val % 10)
        start_node = l3
        carry = (temp_val / 10)
        while l1.next is not None or l2.next is not None:

            v1 = l1.next.val if l1.next is not None else 0
            v2 = l2.next.val if l2.next is not None else 0
            temp_val = v1 + v2 + carry
            new_node = ListNode(temp_val % 10)
            carry = temp_val / 10
            l3.next = new_node
            l3 = new_node

            if l1.next is not None:
                l1 = l1.next
            if l2.next is not None:
                l2 = l2.next

        if carry:
            new_node = ListNode(carry)
            l3.next = new_node
        return start_node

a = Solution()
l1 = ListNode(2)
l1.next = ListNode(4)
# l1.next.next = ListNode(3)

l2 = ListNode(5)
l2.next = ListNode(6)
l2.next.next = ListNode(4)


def print_list(l):
    print(l.val, end=" ")
    while l.next is not None:
        print("->", end=" ")
        print(l.next.val, end=" ")
        l = l.next

print_list(a.addTwoNumbers(l1, l2))