2. 两数相加
2. 两数相加
这道题唯一的难点就是cur如何往后移
题目：
题解：
/** * 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 list = new ListNode(0); ListNode cur = list; int y = 0; while(l1 != null && l2 != null) { cur.next = new ListNode((l1.val + l2.val + y) % 10); y = (l1.val + l2.val + y) / 10; l1 = l1.next; l2 = l2.next; cur = cur.next; } while(l1 != null) { cur.next = new ListNode((l1.val + y) % 10); y = (l1.val + y) / 10; l1 = l1.next; cur = cur.next; } while(l2 != null) { cur.next = new ListNode((l2.val + y) % 10); y = (l2.val + y) / 10; l2 = l2.next; cur = cur.next; } if(y != 0) { cur.next = new ListNode(y); cur = cur.next; } //当前头是虚拟头，所以真正的链表是next之后的 return list.next; } }