leetcodeTOP100(26)两数相加

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。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.

示例 2：

输入：l1 = [0], l2 = [0]
输出：[0]

示例 3：

输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]

存在进位，进位最多为1

双指针法就可以了

package TOP21_30;import Util.ListNode;// 两数相加
public class Top26 {public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return null;}ListNode res = new ListNode(0);ListNode cur = res;int inc = 0;while (l1 != null && l2 != null) {int count = l1.val + l2.val + inc;int resNum = count;if (count >= 10) {resNum = count - 10;inc = 1;}if(count<10){inc =0;}ListNode node = new ListNode(resNum);cur.next = node;cur = cur.next;l1 = l1.next;l2 = l2.next;}// 存在进位while (inc == 1) {if (l1 == null && l2 == null) {ListNode node = new ListNode(1);node.next = null;cur.next = node;return res.next;}if (l1 == null && l2 != null) {int count = l2.val + inc;int resNum = count;if (count >= 10) {resNum = count - 10;inc = 1;} else {inc = 0;}ListNode node = new ListNode(resNum);cur.next = node;cur = cur.next;l2 = l2.next;}if (l1 != null && l2 == null) {int count = l1.val + inc;int resNum = count;if (count >= 10) {resNum = count - 10;inc = 1;} else {inc = 0;}ListNode node = new ListNode(resNum);cur.next = node;cur = cur.next;l1 = l1.next;}}// 不存在进位if (inc == 0) {cur.next = l1 == null ? l2 : l1;}return res.next;}public static void main(String[] args) {int[] num1 = {2, 4,3};int[] num2 = {5,6,4};ListNode list1 = ListNode.setNodes(0, num1);ListNode list2 = ListNode.setNodes(0, num2);ListNode list3 = addTwoNumbers(list1,list2);ListNode.printListData(list3);}
}

harryptter / LeetcodeTop100 · GitCode