LeetCode第二题（Add Two Numbers）

Add Two Numbers

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.

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python代码

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        r = []
        carry = 0
        while l1 and l2:
            num = l1.val + l2.val
            bit = int((num + carry)%10)
            r.append(bit)
            carry = int((num + carry)/10)
            l1 = l1.next
            l2 = l2.next
        if l2:  #取长的链表
            l1 = l2
        while l1:   #处理较长的链表
            num = l1.val + carry
            bit = int((num)%10)
            r.append(bit)
            carry = int((num)/10)
            l1 = l1.next
            if l1 == None:
                if carry != 0:
                    bit = int(carry)
                    r.append(bit)
                carry = 0
                break
        if carry > 0:   #处理多出来的进位
            bit = int(carry)
            r.append(bit)
        return r

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另一种解法

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        head = r = ListNode(0)
        carry = 0
        while l1 and l2:
            num = l1.val + l2.val
            bit = int((num + carry)%10)
            carry = int((num + carry)/10)
            while r.next != None:
                r = r.next
            r.next = ListNode(bit)
            l1 = l1.next
            l2 = l2.next
        if l2:
            l1 = l2
        while l1:
            num = l1.val + carry
            bit = int(num%10)
            while r.next != None:
                r = r.next
            r.next = ListNode(bit)
            carry = int(num/10)
            l1 = l1.next
            if l1 == None:
                if carry != 0:
                    bit = int(carry)
                    r = r.next
                    r.next = ListNode(bit)
                carry = 0
                break
        if carry > 0:
            bit = int(carry)
            r = r.next
            r.next = ListNode(bit)
        r = head.next
        return r

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总结：

• 两种解法区别不大：第一种是用列表存储返回list，第二种是用链表存储起来，直接返回整个链表。
• 个人认为主要考链表操作，不同位数的相加，特别要注意进位