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本篇内容主要讲解“Java 怎么返回首节点”，感兴趣的朋友不妨来看看。本文介绍的方法操作简单快捷，实用性强。下面就让丸趣 TV 小编来带大家学习“Java 怎么返回首节点”吧!
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class AddTwoNumbers { public static void main(String[] args) { ListNode l1 = new ListNode(2);
ListNode l1n1 = new ListNode(4);
ListNode l1n2 = new ListNode(3);
ListNode l2 = new ListNode(5);
ListNode l2n1 = new ListNode(6);
ListNode l2n2 = new ListNode(4);
l1.next = l1n1;
l1n1.next = l1n2;
l2.next = l2n1;
l2n1.next = l2n2;
AddTwoNumbers atn = new AddTwoNumbers();
ListNode pre = atn.addTwoNumbers(l1,l2);
Utils.print(pre);
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry =0;
ListNode newHead = new ListNode(0);// 作为暂存首节点
ListNode p1 = l1, p2 = l2, p3=newHead; //p3 作为操作变量
while(p1 != null || p2 != null){ if(p1 != null){
carry += p1.val;
p1 = p1.next;
}
if(p2 != null){
carry += p2.val;
p2 = p2.next;
}
p3.next = new ListNode(carry%10);
p3 = p3.next;
carry /= 10;
}
if(carry==1)
p3.next=new ListNode(1);
return newHead.next; // 返回首节点
}
}到此，相信大家对“Java 怎么返回首节点”有了更深的了解，不妨来实际操作一番吧！这里是丸趣 TV 网站，更多相关内容可以进入相关频道进行查询，关注我们，继续学习！
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