⭐ 题目日期：

小米 - 2024/11/1

🌳 题目描述：

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]

🕵🏽 面试评估：

这道题考察链表的基础操作，属于简单题

🧗难度系数：

⭐️ ⭐️

📝思路分析：

合并两个有序链表，新链表的头节点不确定，所以我们可以创建一个哑节点 (DummyNode) 作为新链表的头节点，再依次遍历两个链表将较小的节点(比较 Cur 1 与 Cur 2 的大小)连接到新链表，重复以上操作直到所有节点都连接到新链表上，最后返回 DummyNode.next。

💻代码 ：

Java

/**
 * 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 list1, ListNode list2) {
        ListNode dummyNode = new ListNode(0);
        ListNode cur = dummyNode;
        ListNode cur1 = list1;
        ListNode cur2 = list2;

        while (cur1 != null && cur2 != null) {
            if (cur1.val <= cur2.val) {
                cur.next = cur1;
                cur1 = cur1.next;
            } else {
                cur.next = cur2;
                cur2 = cur2.next;
            }

            cur = cur.next;
        }

        if (cur1 == null) {
            cur.next = cur2;
        } else {
            cur.next = cur1;
        }

        return dummyNode.next;
    }
}

Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        dummyNode = ListNode(0)
        cur = dummyNode
        cur1 = list1
        cur2 = list2

        while cur1 is not None and cur2 is not None:
            if cur1.val <= cur2.val:
                cur.next = cur1
                cur1 = cur1.next
            else:
                cur.next = cur2
                cur2 = cur2.next
            cur = cur.next

        if cur1 is None:
            cur.next = cur2
        else:
            cur.next = cur1

        return dummyNode.next

C++

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
        ListNode dummyNode(0);
        ListNode* cur = &dummyNode;
        ListNode* cur1 = list1;
        ListNode* cur2 = list2;

        while (cur1 != nullptr && cur2 != nullptr) {
            if (cur1->val <= cur2->val) {
                cur->next = cur1;
                cur1 = cur1->next;
            } else {
                cur->next = cur2;
                cur2 = cur2->next;
            }
            
            cur = cur->next;
        }

        if (cur1 == nullptr) {
            cur->next = cur2;
        } else {
            cur->next = cur1;
        }

        return dummyNode.next;
    }
};

🚵🏻复杂度分析：

时间复杂度：O(n + m), n, m 分别为两链表的长度

空间复杂度：O(1)

🔬在线评测(Online Judge):

题目链接