LeetCode in Python-2. 两数相加

题目描述

解法 

# 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
        """
        if l1==None:
            return(l2)
        if l2==None:
            return(l1)
        
        carry=0  #满足10向前进位
        dummy=ListNode(0)  #最前面链表的占位
        p=dummy 
        
        while l1 and l2:
            p.next=ListNode((l1.val+l2.val+carry)%10) # 链表的下一位，对10取余的值为下一位
            carry=(l1.val+l2.val+carry)//10 #对10取商，作为向前的进位
            l1=l1.next
            l2=l2.next
            p=p.next
            
        # 当l1与l2位数不同时的情况
        if l1:
            while l1:
                p.next=ListNode((l1.val+carry)%10)
                carry=(l1.val+carry)//10
                l1=l1.next
                p=p.next
        
        if l2:
            while l2:
                p.next=ListNode((l2.val+carry)%10)
                carry=(l2.val+carry)//10
                l2=l2.next
                p=p.next
        
        #当最后一位向前有进位时  
        if carry==1:
            p.next=ListNode(1)
        
        return dummy.next