2. Add Two Numbers
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 contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
Constraints:
- The number of nodes in each linked list is in the range
[1, 100]
. 0 <= Node.val <= 9
- It is guaranteed that the list represents a number that does not have leading zeros.
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 != null
,这样最后一位还需要额外计算,循环终止条件应该是 p != null
。
代码
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function (l1, l2) {
var List = new ListNode(0);
var head = List;
var sum = 0;
var carry = 0;
while (l1 !== null || l2 !== null || sum > 0) {
if (l1 !== null) {
sum = sum + l1.val;
l1 = l1.next;
}
if (l2 !== null) {
sum = sum + l2.val;
l2 = l2.next;
}
if (sum >= 10) {
carry = 1;
sum = sum - 10;
}
head.next = new ListNode(sum);
head = head.next;
sum = carry;
carry = 0;
}
return List.next;
};
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