[LeetCode]2-两数相加

2. 两数相加
给定两个非空链表来表示两个非负整数。位数按照逆序方式存储，它们的每个节点只存储单个数字。将两数相加返回一个新的链表。
你可以假设除了数字 0 之外，这两个数字都不会以零开头。
示例:
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

解法1:

常规思路, 关键点: 头结点, 进位, 链表长度不一致时

class Solution:
    def addTwoNumbers(self, l1, l2):
        carry = 0  #进位值
        isFirst = True  #头结点flag
        currentNode = None  #当前结点
        head = None  #头结点
        while (l1 or l2 or carry != 0): #循环条件: l1, l2的遍历没结束, 或有高位进位时
            val1 = 0
            val2 = 0
            if l1:
                val1 = l1.val
                l1 = l1.next
            if l2:
                val2 = l2.val
                l2 = l2.next
            unit = val1 + val2 + carry
            if unit >= 10:
                carry = 1
                unit = unit%10
            else:
                carry = 0
            
            node =  ListNode(unit)
            if isFirst:
                currentNode = node
                head = currentNode
                isFirst = False
            else:
                currentNode.next = node
                currentNode = currentNode.next
        return head

解法2

给定一个空的头结点, 简化代码.

class Solution:
    def addTwoNumbers(self, l1, l2):
        head = ListNode(0)  #头结点设为空结点
        current = head
        carry = 0
        while (l1 or l2 or carry !=0):
            if l1:
                carry += l1.val
                l1 = l1.next
            if l2:
                carry += l2.val
                l2 = l2.next
            node = ListNode(carry%10)
            carry = carry//10
            current.next = node
            current = current.next

        return head.next