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) {
        if (head1 == null) return head2;
        if (head2 == null) return head1;
        // 根据题目要求，先反转链表
        ListNode new1 = reverseList(head1);
        ListNode new2 = reverseList(head2);
        // 链表相加
        int cnt = 0; // 表示是否有进位
        int temp = 0;
        ListNode p1 = new1;
        ListNode p2 = new2;
        while (p1.next != null && p2.next != null) {
            temp = p1.val + p2.val + cnt;
            p1.val = p2.val = temp % 10;
            cnt = temp / 10;
            p1 = p1.next;
            p2 = p2.next;
        }
        temp = p1.val + p2.val + cnt;
        p1.val = p2.val = temp % 10;
        cnt = temp / 10;
        boolean flag = p2.next == null; // 判断谁最长 false 表示 p2长
        while (p1.next != null) {
            if (cnt == 0) break;
            p1 = p1.next;
            temp = p1.val + cnt;
            p1.val = temp % 10;
            cnt = temp / 10;
        }
        while (p2.next != null) {
            if (cnt == 0) break;
            p2 = p2.next;
            temp = p2.val + cnt;
            p2.val = temp % 10;
            cnt = temp / 10;
        }
        if (cnt != 0) {
            if (flag) {
                // 表示p1 为最终结果
                p1.next = new ListNode(cnt);
                return reverseList(new1);
            } else {
                p2.next = new ListNode(cnt);
                return reverseList(new2);
            }
        } else {
            if (flag) return reverseList(new1);
            return reverseList(new2);
        }
    }
    // 链表反转——头插法
    private ListNode reverseList(ListNode head) {
        if (head == null) return null;
        ListNode pre = null;
        ListNode newHead = head;
        ListNode next;
        while (newHead != null) {
            next = newHead.next;
            newHead.next = pre;
            pre = newHead;
            newHead = next;
        }
        return pre;
    }
}

代码行数有点多，但量不多。主要还是思路比较简单，所以写的判断就有点小多。采用的计算方法也是比较繁琐的，不知道哪位大佬有更加简便的方法，欢迎评论。