给定两个非空链表来表示两个非负整数。位数按照逆序方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。
你可以假设除了数字 0 之外,这两个数字都不会以零开头。
You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
//生成ListNode链表对象,链表的值为0,没有指向的节点
ListNode dummyHead = new ListNode(0);
ListNode p = l1, q = l2, curr = dummyHead;
int carry=0;
while(p!=null||q!=null)
{
int x=(p!=null)?p.val:0;
int y=(q!=null)?q.val:0;
int sum = carry+x+y;
carry=sum/10;
curr.next=new ListNode(sum % 10);
curr=curr.next;
if(p!=null) p=p.next;
if(q!=null) q=q.next;
}
if(carry>0){
//最后一位若还有进位,如果有进位,把进位放到下一个位置里
curr.next=new ListNode(carry);
}
return dummyHead.next;
}
}
错误,首先,创建新的链表的时候没有赋初值,
其次,判断进位的时候要用while循环,上次的进位被这次所覆盖掉
最后,最后还有进位的话,就再放到新的位置里去。
注意:
curr = dummyHead 指 dummyHead 被 curr引用 ,当curr.next改变,那么dummyHead.next也将被改变。 为什么不是返回curr? 因为curr在curr = curr.next已经改变了,而dummyHead是不会因为curr = curr.next而变化的