LeetCode(四)链表206. Reverse Linked List&2. Add Two Numbers

206. Reverse Linked List

翻转一个链表
解法如下

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseList(ListNode head) {        ListNode newhead=null;        while(head!=null){            ListNode next=head.next;            head.next=newhead;            newhead=head;            head=next;        }        return newhead;    }}

2. Add Two Numbers

题目要求

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8

解法如下

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode head1=l1;        ListNode head2=l2;        ListNode res=new ListNode(0);        int sum=0;        ListNode d=res;        while(head1!=null||head2!=null){            sum/=10;            if(head1!=null){                sum+=head1.val;                head1=head1.next;            }            if(head2!=null){                sum+=head2.val;                head2=head2.next;            }            d.next=new ListNode(sum%10);            d=d.next;        }        if(sum/10==1){//进位            d.next=new ListNode(1);        }        return res.next;    }}