2. Add Two Numbers

【题目】

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.

【举例】

Example:

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

【题目翻译】

给定两个数字链表， 然后把它们相加， 得到和链表， 具体见上面的example。

【解题思路】

其实这道题思路还是挺明显的， 无非就是同步遍历两个链表， 相加求和。

【实现代码】

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        lead = ListNode(0)
        p = lead
        carry = 0
        while l1 != None and l2 != None: #两个列表同步遍历， 依次相加
            value = (carry+ l1.val + l2.val)%10
            p.next = ListNode(value)
            p = p.next
            carry = (carry + l1.val + l2.val)//10
            l1 = l1.next
            l2 = l2.next
        remain = l1 if l1 != None else l2 #把剩余的一个列表加完
        while remain != None: 
            value = (carry + remain.val)%10
            p.next = ListNode(value)
            p = p.next
            carry = (carry + remain.val)//10
            remain = remain.next
        if carry == 1: #如果列表加完以后还有进位
            p.next = ListNode(1)
        return lead.next