You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8
题目解释：两个链表对应位置上的数字：2 + 5 = 7. 4 + 6 = 10 进位1 . 4 + 3 + 1（进位） = 8.
思路：同时遍历两个链表，将它们对应位置上的值相加，并以此值新建节点，注意进位情况。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { if(l1 == null) return l2; if(l2 == null) return l1; ListNode dummy = new ListNode(0); ListNode head = dummy; ListNode head1 = l1, head2 = l2; int c = 0 ,sum = 0; //同时遍历链表l1和链表l2 while(head1 != null && head2 !=null){ sum = head1.val + head2.val + c; //进位量 c = sum / 10; head.next = new ListNode(sum % 10); head1 = head1.next; head2 = head2.next; head = head.next; } //以下两个while循环只会执行其中一个 //若链表l1未遍历完，则继续遍历l1(此情况是链表l1的长度大于链表l2) while(head1 != null){ sum = head1.val + c; c = sum / 10; head.next = new ListNode(sum % 10); head1 = head1.next; head = head.next; } //若链表l2未遍历完，则继续遍历l2(此情况是链表l2的长度大于链表l1) while(head2 != null){ sum = head2.val + c; c = sum / 10; head.next = new ListNode(sum % 10); head2 = head2.next; head = head.next; } //若最后有进位，则新建一个节点，也是最后一个节点 if(c != 0){ head.next = new ListNode(c); } return dummy.next; } }
