题解 | #链表相加(二)#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        // write code here
        head1 = ReverseList(head1);
        head2 = ReverseList(head2);
        
        ListNode t1 = head1, t2 = head2;
        
        ListNode result = null; // 头指针
        ListNode res_tail = null;
        
        int cin = 0;
        
        while (t1 != null || t2 != null) {
            
            int count = 0;
            if (t1 != null) {
                count += t1.val;
                t1 = t1.next;
            }
            if (t2 != null) {
                count += t2.val;
                t2 = t2.next;
            }
            
            count += cin;
            
            ListNode t = new ListNode(count % 10);
            cin = count / 10;
            
            if (result == null) {
                result = t;
                res_tail = result;
            }
            else {
                res_tail.next = t;
                res_tail = res_tail.next;
            }
        }
        
        if (cin > 0)
             res_tail.next = new ListNode(cin);
        
        return ReverseList(result);
    }
    
    public ListNode ReverseList(ListNode head) {
        if (head == null || head.next == null){
            return head;
        }
        ListNode r = head.next;
        ListNode l = head;
        ListNode t = r.next;
        head.next = null;
        
        while(t != null) {
            r.next = l;
            l = r;
            r = t;
            t = r.next;
        }
        r.next = l;
        
        return r;
    }
}

主要是链表先反转，直接copy的前几个题的，反转后for循环长链做加法，记录进位cin，最后反转输出列表