JZ-016-Merge two sorted linked lists

Merge two sorted linked lists

Title description

Input two monotonically increasing linked lists, and output the synthesized linked list of the two linked lists. Of course, we need the synthesized linked list to satisfy the monotonous non-decreasing rule.

title link : merges two sorted linked lists

Code

/**
 * 标题：合并两个排序的链表
 * 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。
 * 题目链接：
 * https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337?tpId=13&&tqId=11169&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
 */
public class Jz16 {
  
    public ListNode merge(ListNode list1, ListNode list2) {
        if (list1 == null && list2 == null) {
            return null;
        }
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }
        ListNode result = new ListNode(-1);
        ListNode next = result;
        while (list1 != null || list2 != null) {
            if (list1 == null) {
                next.next = list2;
                break;
            } else if (list2 == null) {
                next.next = list1;
                break;
            } else if (list1.val < list2.val) {
                next.next = list1;
                list1 = list1.next;
            } else {
                next.next = list2;
                list2 = list2.next;
            }
            next = next.next;
        }

        return result.next;
    }

    public static void main(String[] args) {
        ListNode list1 = new ListNode(1);
        list1.next = new ListNode(3);
        list1.next.next = new ListNode(5);

        ListNode list2 = new ListNode(2);
        list2.next = new ListNode(4);
        list2.next.next = new ListNode(6);

        System.out.println("after merge");
        Jz16 jz16 = new Jz16();
        ListNode result = jz16.merge(list1, list2);
        ListNode cur = result;
        while (cur != null) {
            System.out.print(cur.val + " ");
            cur = cur.next;
        }
    }
}

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