面试经典-合并两个有序链表

合并两个有序链表

题目

21. 合并两个有序链表

将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示：

• 两个链表的节点数目范围是 [0, 50]
• -100 <= Node.val <= 100
• l1 和 l2 均按 非递减顺序 排列

题解

简单，自己做

public class MergeTwoLists {

    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(2);
        l1.next.next = new ListNode(4);
        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(3);
        l2.next.next = new ListNode(4);
        ListNode node = mergeTwoLists(l1, l2);
        while (node != null) {
            System.out.println(node.val);
            node = node.next;
        }
    }

    public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode result = null;
        ListNode item = null;
        if (l1 == null || l2 == null) {
            return l1 == null ? l2 : l1;
        }
        while (true) {
            if (l1 == null) {
                item.next = l2;
                return result;
            }
            if (l2 == null) {
                item.next = l1;
                return result;
            }
            int l1val = l1.val;
            int l2val = l2.val;
            if (l1val <= l2val) {
                if (result == null) {
                    result = item = new ListNode(l1val);
                }else {
                    item.next = new ListNode(l1val);
                    item = item.next;
                }
                l1 = l1.next;
            }else {
                if (result == null) {
                    result = item = new ListNode(l2val);
                }else {
                    item.next = new ListNode(l2val);
                    item = item.next;
                }
                l2 = l2.next;
            }
        }
    }

}

class ListNode {
    int val;
    ListNode next;

    ListNode() {}

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

    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}