Description

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.

Solution

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        carry_bit = 0 # 进位值
        cur_bit = 0 # 当前位数值
        prev_node = ListNode(0)
        start_node = prev_node
        cur_node_1 = l1
        cur_node_2 = l2
        while True:
            if cur_node_1 and cur_node_2: # 如果两个链表当前节点均不为空
                prev_node.next = ListNode(0)
                val = cur_node_1.val + cur_node_2.val + carry_bit
                cur_bit = val % 10
                carry_bit = val // 10
                # print(cur_node_1.val, cur_node_2.val, cur_bit, carry_bit)

                prev_node.next.val = cur_bit
                prev_node = prev_node.next

                # print_linked_list(start_node)
            else:
                break

            cur_node_1 = cur_node_1.next # 指向链表下一个节点
            cur_node_2 = cur_node_2.next # 指向链表下一个节点

        if cur_node_1 is None and cur_node_2 is None and carry_bit < 1:
            # print_linked_list(start_node)
            pass
        elif cur_node_1:
            while True:
                val = cur_node_1.val + carry_bit
                cur_bit = val % 10
                carry_bit = val // 10

                prev_node.next = ListNode(0)
                prev_node.next.val = cur_bit
                prev_node = prev_node.next

                cur_node_1 = cur_node_1.next
                if cur_node_1 is None:
                    break
        elif cur_node_2:
            while True:
                val = cur_node_2.val + carry_bit
                cur_bit = val % 10
                carry_bit = val // 10

                prev_node.next = ListNode(0)
                prev_node.next.val = cur_bit
                prev_node = prev_node.next

                cur_node_2 = cur_node_2.next
                if cur_node_2 is None:
                    break
        if carry_bit > 0:
            prev_node.next = ListNode(0)
            prev_node.next.val = carry_bit
            prev_node = prev_node.next

        return start_node.next

辅助代码

def print_linked_list(l1):
    # 打印链表
    while True:
        print(l1.val, end=' ')
        l1 = l1.next
        if l1 is None:
            break
    print()

备注

该问题旨在利用链表数据结构模拟带进位加法。解题过程注意进位及最高位各种情形的处理。