剑指offer---合并两个排序的链表-优快云博客

本文链接：https://blog.youkuaiyun.com/dzq_feixiang/article/details/51747205

这篇博客介绍了如何合并两个单调递增的链表，确保合并后的链表依然保持单调不减的顺序。文章提供了两种解法，并给出了完整的代码实现。

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题目描述
输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

解法一：

public static ListNode Merge(ListNode list1, ListNode list2) {
        if (list1 == null)
            return list2;
        if (list2 == null)
            return list1;
        // 创建一个临时结点，用于添加元素时方便
        ListNode root = new ListNode(0);
        // 用于指向合并后的新链的尾结点
        ListNode pointer = root;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                pointer.next = list1;
                list1 = list1.next;
            } else {
                pointer.next = list2;
                list2 = list2.next;
            }
            pointer = pointer.next;
        }

        if (list1 != null) {
            pointer.next = list1;
        }
        if (list2 != null) {
            pointer.next = list2;
        }
        return root.next;
    }

解法二：

public static ListNode Merge2(ListNode list1, ListNode list2) {
        if (list1 == null)
            return list2;
        if (list2 == null)
            return list1;
        ListNode head = null;
        if (list1.val <= list2.val) {
            head = list1;
            head.next = Merge2(list1.next, list2);
        } else {
            head = list2;
            head.next = Merge2(list1, list2.next);
        }
        return head;
    }

完整代码：

public class ListNode {

    int val;
    ListNode next = null;

    public ListNode(int val) {
        this.val = val;
    }

    public static void printList(ListNode head) {
        while (head != null) {
            System.out.print(head.val + "->");
            head = head.next;
        }
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);

        printList(head);
        System.out.println();
        ListNode head2 = new ListNode(1);
        head2.next = new ListNode(3);
        head2.next.next = new ListNode(5);
        head2.next.next.next = new ListNode(6);
        head2.next.next.next.next = new ListNode(7);
        printList(head2);
        System.out.println();

        head = Merge(head, head2);
        // head = Merge2(head, head2);
        printList(head);
    }
}