LeetCode 线性表篇：Add Two Number

一、题目描述

题目原文：

 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.

两个单链表分别表示两个非负整数，表示形式为：数字个位—>数字十位—>数字百位........请以相同的单链表的形式返回两个数的和的结果

Example：

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

二、解题思路

此题不必按照咱们平时的早已根深蒂固的加法法则，这样考虑的话比较麻烦，就可以简单地看做两个单链表 node 相加，但是与加法不同的是从百位开始，遇到进位向右进位。

三、代码（来自//blog.csdn.net/lanchunhui/article/details/51119205）

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

# 为链表添加新的元素
def addTail(l, node):
    if l.next is None:
        l.next = node
    else:
        p = l.next
        while p.next:
            p = p.next
        p.next = node

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        carrier = 0
        x, y = l1.val, l2.val
        if x+y>=10:
            carrier = 1
            l = ListNode(x+y-10)
        else:
            l = ListNode(x+y)
        while l1.next and l2.next:
            x, y = l1.next.val, l2.next.val
            if x+y+carrier>=10:
                addTail(l, ListNode(x+y+carrier-10))
                carrier = 1
            else:
                addTail(l, ListNode(x+y+carrier))
                carrier = 0
            l1 = l1.next
            l2 = l2.next

        # 以下两个 while 循环处理长度不一致的情况
        while l1.next:
            if l1.next.val+carrier >= 10:
                addTail(l, ListNode(l1.next.val+carrier-10))
                carrier = 1
            else:
                addTail(l, ListNode(l1.next.val+carrier))
                carrier = 0
            l1 = l1.next
        while l2.next:
            if l2.next.val+carrier >= 10:
                addTail(l, ListNode(l2.next.val+carrier-10))
                carrier = 1
            else:
                addTail(l, ListNode(l2.next.val+carrier))
                carrier = 0
            l2 = l2.next

        # 还有进位
        if carrier == 1:
            addTail(l, ListNode(1))
        return l

if __name__ == '__main__':
    print("----------------- start -----------------")

    l1_1 = ListNode(2)
    l1_2 = ListNode(4)
    l1_3 = ListNode(3)
    l1_1.next = l1_2
    l1_2.next = l1_3

    l2_1 = ListNode(5)
    l2_2 = ListNode(6)
    l2_3 = ListNode(4)
    l2_1.next = l2_2
    l2_2.next = l2_3

    l3_1 = Solution().addTwoNumbers(l1_1, l2_1)
    while l3_1 != None:
        print(l3_1.val)
        l3_1 = l3_1.next