两个链表生成相加链表

题目描述

假设链表中每一个节点的值都在 0 - 9 之间,那么链表整体就可以代表一个整数。

给定两个这种链表,请生成代表两个整数相加值的结果链表。

链表 1 为 9->3->7,链表 2 为 6->3,最后生成新的结果链表为 1->0->0->0。

示例

输入:(6 -> 1 -> 7) + (2 -> 9 -> 5),即617 + 295
输出:9 -> 1 -> 2,即912

解决方法

  1. 反转两个链表
  2. 同时遍历两个链表,依次相加生成新节点,并记录进位
  3. 反转新链表

==> 还可以简单进阶一下,不完全生成一条全新的链表。也就是说在任选一条链表基础上赋值结果,若判断少了节点,再新建,并直接连接在该链表后

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 || head2 == null) {
     
            return head1 == null ? head2 : head1;
        }
        head1 = reverse(head1);
        head2 = reverse(head2);
        ListNode ptail = new ListNode(-1);
        ListNode phead = ptail, pnew = ptail;
        int temp = 0;            //进位
        while (head1 != null || head2 != null || temp > 0) {
     
            int num1 = (head1 == null ? 0 : head1.val);
            int num2 = (head2 == null ? 0 : head2.val);
            int num = num1 + num2 + temp;
            pnew = new ListNode(num%10);
            ptail.next = pnew;
            ptail = ptail.next;
            temp = num/10;
            head1 = (head1 == null ? null : head1.next);
            head2 = (head2 == null ? null : head2.next);
        }
        phead = reverse(phead.next);
        return phead;
    }
    
    public ListNode reverse (ListNode head) {
     
        if (head == null || head.next == null) {
     
            return head;
        }
        ListNode preNode = head;
        ListNode target = head.next;
        while (target != null) {
     
            preNode.next = target.next;
            target.next = head;
            head = target;
            target = preNode.next;
        }
        return head;
    }
}

进阶

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 || head2 == null) {
     
            return head1 == null ? head2 : head1;
        }
        ListNode ptail = head2;            //定位尾结点
        head1 = reverse(head1);
        head2 = reverse(head2);
        ListNode phead = head2;
        int temp = 0;            //进位
        while (head1 != null || head2 != null || temp > 0) {
     
            int num1 = (head1 == null ? 0 : head1.val);
            int num2 = (head2 == null ? 0 : head2.val);
            int num = num1 + num2 + temp;
            if (head2 != null) {
     
                head2.val = num%10;
            } else {
     
                ptail.next = new ListNode(num%10);
                ptail = ptail.next;
            }
            temp = num/10;
            head1 = (head1 == null ? null : head1.next);
            head2 = (head2 == null ? null : head2.next);
        }
        phead = reverse(phead);
        return phead;
    }
    
    public ListNode reverse (ListNode head) {
     
        if (head == null || head.next == null) {
     
            return head;
        }
        ListNode preNode = head;
        ListNode target = head.next;
        while (target != null) {
     
            preNode.next = target.next;
            target.next = head;
            head = target;
            target = preNode.next;
        }
        return head;
    }
}

你可能感兴趣的:(算法题,链表,java,算法)