出處： https://leetcode.com/problems/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 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.
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
let temp = 0;
const head = new ListNode()
let node = head
while(l1 || l2) {
let sum = temp
if (l1 !== null) {
sum += l1.val
l1 = l1.next
}
if (l2 !== null) {
sum += l2.val
l2 = l2.next
}
if (sum >= 10) {
temp = 1
sum = sum - 10
} else {
temp = 0
}
node.next = new c(sum)
node = node.next
}
if (temp === 1) {
node.next = new ListNode(1)
}
return head.next
};思路
這題如果做過基本的 Linked List 題目，應該不難。建議先把以下這幾題搞懂：
- 解題 30. Design Linked List (節點鏈)
- 解題 31. Reverse Linked List (反轉 Linked List)
- 解題 35. Palindrome Linked List
首先，這邊 input 的 linked list 已經是從個位數開始往下連結。所以不需要知道整個 list 的全貌即可開始處理。
將兩個 l1 和 l2 的 val 相加，由於題目已經規範數字是在 1 ~ 9 之間，所以兩數相加最大值不過 18 (兩個 val 都是 9)，因此如果兩數的和 sum 超過 10，則把 1 帶入下一個階段。
等到 l1, l2 都已經跑完變成 null ，在檢查最後一次相加是否超過 10 。如果超過，要再補一個 val 為 1 的 ListNode
