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
        ListNode l1 = reverse(head1);
        ListNode l2 = reverse(head2);
        ListNode result = null;
        int flag = 0;
        int i = 0;
	  //考虑（6，3）（1，1）这种情况
        while (l1 != null && l2 != null) {
            ListNode temp = new ListNode((l1.val + l2.val + flag) % 10);
            flag = (l1.val + l2.val + flag) / 10;
            temp.next = result;
            result = temp;
            l1 = l1.next;
            l2 = l2.next;
            System.out.print(++i);
            System.out.println(result.val);
        }
	  //考虑（6，3）（1）这种情况
        if (l2 == null) {
            while (l1 != null) {
                ListNode temp = new ListNode((l1.val  + flag) % 10);
                flag = (l1.val + flag) / 10;
                temp.next = result;
                result = temp;
                l1 = l1.next;
            }
        }
	  //考虑（1）（6，3）这种情况
        if (l1 == null) {
            while (l2 != null) {
                ListNode temp = new ListNode((l2.val + flag) % 10);
                flag = (l2.val + flag) / 10;
                temp.next = result;
                result = temp;
                l2 = l2.next;
            }
        }
	  //考虑最后结果为1000这种的。（1是进位）
        if (flag == 1) {
            ListNode temp = new ListNode(flag);
            temp.next = result;
            result = temp;
        }
        return result;
    }
//头插法逆序
    public ListNode reverse(ListNode head) {
        ListNode q = null;
        ListNode p = head;
        while (p != null) {
            ListNode Temp = p.next;
            p.next = q;
            q = p;
            p = Temp;
        }
        return q;
    }
}