关于java:JZ016合并两个排序的链表

40次阅读

共计 1191 个字符，预计需要花费 3 分钟才能阅读完成。

输出两个枯燥递增的链表，输入两个链表合成后的链表，当然咱们须要合成后的链表满足枯燥不减规定。

题目链接 : 合并两个排序的链表

/**
 * 题目：合并两个排序的链表
 * 题目形容
 * 输出两个枯燥递增的链表，输入两个链表合成后的链表，当然咱们须要合成后的链表满足枯燥不减规定。* 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;
        }
    }
}