LintCode Python 简单级题目 167.链表求和

原题描述：

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

您在真实的面试中是否遇到过这个题？ 

Yes

样例

给出两个链表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null

标签 

Cracking The Coding Interview 链表 高精度

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

class Solution:
    # @param l1: the first list
    # @param l2: the second list
    # @return: the sum list of l1 and l2 
    def addLists(self, l1, l2):
        # write your code here

　　

 

题目分析：

函数addLists，入参为两个链表，长度无限制，即可能存在长度：list1 > list2; list1 = list2; list1 < list2;

最终链表长度依据最长链表长度n，返回链表长度（n~n+1）

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

class Solution:
    # @param l1: the first list
    # @param l2: the second list
    # @return: the sum list of l1 and l2 
    def addLists(self, l1, l2):
        # write your code here
        if l1 is None: return l2
        if l2 is None: return l1
        
        head1 = l1
        head2 = l2
        flag = 0
        
        while head1.next is not None or head2.next is not None:
            # 存在某一链表next为空时，构造next.val = 0，不影响加法结果
            if head1.next is None:
                head1.next = ListNode(0)
            if head2.next is None:
                head2.next = ListNode(0)
                
            sumNum = head1.val + head2.val
            if sumNum >= 10:
                head1.val = sumNum%10
                flag = 1
                head1.next.val += 1
            else:
                head1.val = sumNum
                flag = 0
            head1 = head1.next
            head2 = head2.next
        else:
            # 链表末尾时，单独处理，其和大于10时，追加节点
            head1.val = head1.val + head2.val
            if head1.val >= 10:
                head1.val = head1.val%10
                head1.next = ListNode(1)
        return l1