【LeetCode】Add the linked list with carry

/*** Definition for singly-linked list.* public class ListNode {* int val;*ListNode next;* ListNode() {}* ListNode(int val) { this.val = val; }* ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode pre = new ListNode(0); //Open up a new memory space, pre points to the memory space, and pre is the head node of the result listListNode cur = pre; //Define a movable pointer that also points to this memory space, and subsequent addition of nodes depends on this pointerint carry = 0; //carrywhile(l1!=null || l2!=null){int x = l1 != null ? l1.val : 0; //Get the value at the corresponding position, if a position is empty, assign it to 0int y = l2 != null ? l2.val : 0;int sum = x + y + carry; //Add the value at the corresponding position of l1 and l2 and the value of the carrycarry = sum / 10; //Get the carry after each additionsum = sum % 10; //Get the value after the carrycur.next = new ListNode(sum); //Create a new node, the value of the node is the value after adding the carry, and the pointer of the resulting linked list points to the nodecur = cur.next; //The result linked list pointer moves forward, the pre head pointer does not moveif(l1!= null){l1 = l1.next;}if(l2!= null){l2 = l2.next;}}if(carry == 1){cur.next = new ListNode(carry); //If there is a carry after the last addition (the carry value must be 1), create a new node and save the node}return pre.next;//Return the result linked list, because pre points to the head pointer of the linked list, the value at the head pointer is 0, and the next node of the head pointer is the first node of the result linked list}}

Open up a new linked list space ListNode pre = new ListNode(0) pre is the head node pointer

The last return is return pre.next, pre.next points to the first node of the result list

Do not use the head node, directly return the result list method