2. Add Two Numbers #

题目 #

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.

题目大意 #

2 个逆序的链表，要求从低位开始相加，得出结果也逆序输出，返回值是逆序结果链表的头结点。

解题思路 #

需要注意的是各种进位问题。

极端情况，例如

Input: (9 -> 9 -> 9 -> 9 -> 9) + (1 -> )
Output: 0 -> 0 -> 0 -> 0 -> 0 -> 1

为了处理方法统一，可以先建立一个虚拟头结点，这个虚拟头结点的 Next 指向真正的 head，这样 head 不需要单独处理，直接 while 循环即可。另外判断循环终止的条件不用是 p.Next ！= nil，这样最后一位还需要额外计算，循环终止条件应该是 p != nil。

代码 #

package leetcode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	head := &ListNode{Val: 0}
	n1, n2, carry, current := 0, 0, 0, head
	for l1 != nil || l2 != nil || carry != 0 {
		if l1 == nil {
			n1 = 0
		} else {
			n1 = l1.Val
			l1 = l1.Next
		}
		if l2 == nil {
			n2 = 0
		} else {
			n2 = l2.Val
			l2 = l2.Next
		}
		current.Next = &ListNode{Val: (n1 + n2 + carry) % 10}
		current = current.Next
		carry = (n1 + n2 + carry) / 10
	}
	return head.Next
}