Leetcode-21-合并两个有序链表

21. 合并两个有序链表

难度简单1687收藏分享切换为英文接收动态反馈

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：

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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

21. Merge Two Sorted Lists

Easy

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Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists.

Example 1:

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Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: l1 = [], l2 = []
Output: []

Example 3:

Input: l1 = [], l2 = [0]
Output: [0]

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both l1 and l2 are sorted in non-decreasing order.

使用递归实现，新链表也不需要构造新节点

• 终止条件：两条链表分别名为 l1 和 l2，当 l1 为空或 l2 为空时结束
• 返回值：每一层调用都返回小的链表头，然后返回节点的next节点继续参与递归

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) {//终止条件
            return l2;
        }
        if(l2 == null) {//终止条件
            return l1;
        }

        if(l1.val < l2.val) {//链表头比较大小，谁小谁作为返回节点
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

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