LintCode 221 [Add Two Numbers II]

原题

假定用一个链表表示两个数，其中每个节点仅包含一个数字。假设这两个数的数字顺序排列，请设计一种方法将两个数相加，并将其结果表现为链表的形式。

样例
给出 6->1->7 + 2->9->5。即，617 + 295。
返回 9->1->2。即，912 。

解题思路

• 整体思路与Add Two Numbers一样，只不过链表和其表示的数值吮吸相同
• 借助helper函数，反转链表

完整代码

# 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 addLists2(self, l1, l2):
        # Write your code here
        l1 = self.reverse(l1)
        l2 = self.reverse(l2)
        carry = 0
        Dummy = ListNode(0)
        head = Dummy
        while l1 != None and l2 != None:
            sum = (l1.val + l2.val + carry) % 10
            carry = 1 if (l1.val + l2.val + carry) > 9 else 0
            head.next = ListNode(sum)
            head = head.next
            l1 = l1.next
            l2 = l2.next
            
        while l1 != None:
            sum = (l1.val + carry) % 10
            carry = 1 if l1.val + carry > 9 else 0
            head.next = ListNode(sum)
            l1 = l1.next
            head = head.next
        
        while l2 != None:
            sum = (l2.val + carry) % 10
            carry = 1 if l2.val + carry > 9 else 0
            head.next = ListNode(sum)
            l2 = l2.next
            head = head.next
        
        if carry:
            head.next = ListNode(carry)
        
        return self.reverse(Dummy.next)
        
        
    def reverse(self, head):
        if not head:
            return head
            
        prev = None
        while head != None:
            temp = head.next
            head.next = prev
            prev = head
            head = temp
        return prev