91 lines
2.4 KiB
JavaScript
91 lines
2.4 KiB
JavaScript
/* leetcode 2
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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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// 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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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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const x = l1 !== null ? l1.val : 0;
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const y = l2 !== null ? l2.val : 0;
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const 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 /= 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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const l1 = LinkList.fromArray(
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[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],
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).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)));
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console.log(LinkList.prototype.toString.apply(addTwoNumbers(l1, l2)));
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