90 lines
2.3 KiB
JavaScript
90 lines
2.3 KiB
JavaScript
/*
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给你两个 非空 的链表，表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的，并且每个节点只能存储 一位 数字。
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请你将两个数相加，并以相同形式返回一个表示和的链表。
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你可以假设除了数字 0 之外，这两个数都不会以 0 开头。
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输入：l1 = [2,4,3], l2 = [5,6,4]
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输出：[7,0,8]
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解释：342 + 465 = 807.
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示例 2：
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输入：l1 = [0], l2 = [0]
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输出：[0]
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示例 3：
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输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
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输出：[8,9,9,9,0,0,0,1]
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解释：也就是用列表表示
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*/
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function addTwoNumbers(l1, l2) {
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const head = new ListNode()
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let curNode = head
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let carry = 0
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while (l1!==null || l2!==null) {
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let x = l1 !== null ? l1.val : 0
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let y = l2 !== null ? l2.val : 0
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let sum = x + y + carry
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curNode.val = sum % 10
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carry = Math.floor(sum/10)
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if(l1!==null) l1=l1.next
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if(l2!==null) l2=l2.next
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if(carry > 0 || l1 !=null || l2 != null){
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curNode.next = new ListNode()
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curNode = curNode.next
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}
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}
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if(carry!==0) {
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curNode.val = carry
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}
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return head
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}
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function addTwoNumbers2(l1, l2){
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const task1 = []
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const task2 = []
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while(l1 || l2) {
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if(l1){
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task1.push(l1.val)
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l1 = l1.next
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}
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if(l2){
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task2.push(l2.val)
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l2 = l2.next
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}
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}
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let target = BigInt(task1.reverse().join('')) + BigInt(task2.reverse().join(''))
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// console.log(Number(task1.reverse().join('')));
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console.log(target);
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const head = new ListNode()
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let cur = head
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while (target) {
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cur.val = target % 10n
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target = target / 10n
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cur.next = target ? new ListNode() : null
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cur = cur.next
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}
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return head
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}
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// 换一种思路，我们按照顺序压入栈中，再依次弹出，之后生成对应的数，最后再把这个数转化为链表不就行了吗？
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// 输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
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// 输出：[8,9,9,9,0,0,0,1]
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import {LinkList, LinkListNode as ListNode} from '../../list/index.js'
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const l1 = LinkList.fromArray([1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1]).head
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const l2 = LinkList.fromArray([5,6,4]).head
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// console.log(LinkList.prototype.toString.apply(addTwoNumbers(l1,l2)));
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console.log(LinkList.prototype.toString.apply(addTwoNumbers2(l1,l2))); |