剑指offer25 合并两个有序链表 Java

合并两个有序链表

最新推荐文章于 2023-07-26 17:54:13 发布

springTen

最新推荐文章于 2023-07-26 17:54:13 发布

阅读量118

点赞数

CC 4.0 BY-SA版权

分类专栏：剑指offer 算法 Java 链表

本文链接：https://blog.youkuaiyun.com/weixin_43065507/article/details/99332264

算法同时被 3 个专栏收录

47 篇文章

订阅专栏

Java

47 篇文章

订阅专栏

剑指offer

39 篇文章

订阅专栏

本文介绍了一种合并两个已排序的链表的方法，通过遍历两个链表并比较节点值来创建一个新的已排序链表。文章提供了完整的Java代码实现，展示了如何有效地合并链表，并保持元素的排序。

public class MergeSortedLinkList25 {
    static class Node {
        int val;
        Node next;
        public Node(int val){
            this.val = val;
        }
        @Override
        public String toString() {
            return "Node{" +
                    "val=" + val +
                    ", next=" + next +
                    '}';
        }
    }
    public static void main(String[] args) {
        Node root = new Node(0);
        Node node1 = new Node(2);
        Node node2 = new Node(4);
        Node node3 = new Node(6);
        Node node4 = new Node(8);
        Node node5 = new Node(10);
        root.next = node1;
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        node4.next = node5;

        Node root2 = new Node(1);
        Node node21 = new Node(3);
        Node node22 = new Node(5);
        Node node23 = new Node(7);
        Node node24 = new Node(9);
        root2.next = node21;
        node21.next = node22;
        node22.next = node23;
        node23.next = node24;
        System.out.println(merge(root, root2));
    }

    private static Node merge(Node root1, Node root2) {
        Node p1 = root1;
        Node p2 = root2;
        Node head = new Node(-1);
        Node p = head;
        while (p1 != null && p2 != null) {
            Node node = null;
            if (p1.val < p2.val) {
                node = new Node(p1.val);
                p1 = p1.next;
            } else {
                node = new Node(p2.val);
                p2 = p2.next;
            }
            p.next = node;
            p = p.next;
        }
        while (p1 != null) {
            Node node = new Node(p1.val);
            p1 = p1.next;
            p.next = node;
            p = p.next;
        }
        while (p2 != null) {
            Node node = new Node(p2.val);
            p2 = p2.next;
            p.next = node;
            p = p.next;
        }
        p.next = null;
        return head.next;
    }
}