[LeetCode]2-两数相加

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