[剑指offer刷题] Acwing 36 合并两个排序的链表（双指针）-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_22473333/article/details/121776495

这篇博客介绍了两种合并两个已排序链表的方法。第一种使用了双指针，分别遍历两个链表并比较节点值，将较小的节点添加到结果链表中。第二种方法利用虚拟节点，创建一个头节点，然后同样比较节点值并连接到结果链表。两种方法都实现了O(n)的时间复杂度，提高了效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

36 合并两个排序的链表

思路

单纯的双指针，容易想，简单

java代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode merge(ListNode l1, ListNode l2) {

        if (l1 == null && l2 == null) return null;

        if (l1 == null && l2 != null) return l2;
        if (l1 != null && l2 == null) return l1;

        ListNode t1 = l1, t2 = l2;
        ListNode res = null;
        if (t1.val < t2.val) {
            res = t1;
            t1 = t1.next;
        }
        else {
            res = t2;
            t2 = t2.next;   
        }

        ListNode t0 = res;

        while (t1 != null && t2 != null) {
            if (t1.val < t2.val) {
                res.next = t1;
                t1 = t1.next; 
                res = res.next;
            } else {
                res.next = t2;
                t2 = t2.next;
                res = res.next;
            }
        }

        while (t1 != null) {
            res.next = t1;
            t1 = t1.next;
            res = res.next;
        }

        while (t2 != null) {
            res.next = t2;
            t2 = t2.next;
            res = res.next;
        }
        return t0;
    }
}

使用虚拟节点，代码更短

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode merge(ListNode l1, ListNode l2) {

        ListNode dummy = new ListNode(-1);
        ListNode t1 = l1, t2 = l2;
        ListNode cur = dummy;

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next; 
                cur = cur.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
                cur = cur.next;
            }
        }

        if (l1 != null) cur.next = l1;
        else cur.next = l2;
        return dummy.next;
    }
}