合并两个有序链表

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

方法一：迭代

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;

        // 头结点
        ListNode head;
        if (l1.val <= l2.val) {
            head = l1;
            l1 = l1.next;
        } else {
            head = l2;
            l2 = l2.next;
        }

        ListNode cur = head;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                cur = cur.next = l1;
                l1 = l1.next;
            } else {
                cur = cur.next = l2;
                l2 = l2.next;
            }
        }

        if (l1 == null) {
            cur.next = l2;
        } else if (l2 == null) {
            cur.next = l1;
        }
        return head;
    }
}

 

方法二：利用虚拟头结点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;

        // 虚拟头结点
        ListNode head = new ListNode(0);
        ListNode cur = head;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                cur = cur.next = l1;
                l1 = l1.next;
            } else {
                cur = cur.next = l2;
                l2 = l2.next;
            }
        }

        if (l1 == null) {
            cur.next = l2;
        } else if (l2 == null) {
            cur.next = l1;
        }
        return head.next;
    }
}

 

方法三：递归

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;

        if (l1.val <= l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}