2. Add Two Numbers

2. Add Two Numbers

1. 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.

Example:

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

2.思路

本题其实没有什么巧妙的算法。主要是两个链表的数字相加,注意边界条件以及进位就好了

3. 实现

# Definition for singly-linked list.
# 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
        """
        jinw = 0
        
        r = ListNode(0)
        ret = r 
        while True:
            
            if l1 != None and l2 != None:
                ret.val = l1.val + l2.val + jinw
            elif l1 == None and l2 != None:
                ret.val = l2.val + jinw
            elif l2 == None and l1 != None:
                ret.val = l1.val + jinw
            elif jinw != 0 and l2 == None and l1 == None:
                ret.val = jinw
                
            
            if ret.val >= 10 :
                jinw = ret.val /10
                ret.val = ret.val % 10
            else:
                jinw = 0
            
         
            if ( l1 == None and l2 == None and jinw == 0 ):
                break
            if l1 is not None or jinw != 0  : 
                if l1 is None :
                    pass
                else:
                    l1 = l1.next
                
            if l2 is not None or  jinw != 0   :
                if l2 is None :
                    pass
                else:
                    l2 = l2.next
                
            if jinw != 0 or l1 != None or l2 !=None :
                ret.next = ListNode(0)
                ret = ret.next 
               
            
        return r
        
posted @ 2019-05-26 19:32  bush2582  阅读(64)  评论(0编辑  收藏  举报