21. 合并两个有序链表

原创已于 2025-08-17 11:46:15 修改 · 285 阅读

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#链表 #数据结构 #合并两个有序链表

于 2023-12-11 10:38:01 首次发布

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文章介绍了两种方法合并两个链表，法1通过使用虚拟头节点简化代码结构，法2虽然代码更简洁但逻辑清晰。

题目

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]:
        dummy = ListNode(-1)
        cur = dummy
        l1, l2 = list1, list2
        while l1 and l2:
            if l1.val <= l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        cur.next = l1 if l1 else l2
        return dummy.next

Java

法1：最佳解法，使用虚拟头结点

class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode(-1); // 虚拟头结点
        ListNode tmp = dummy;
        while (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                tmp.next = list1;
                list1 = list1.next;
            } else {
                tmp.next = list2;
                list2 = list2.next;
            }
            tmp = tmp.next;
        }
        if (list1 == null) {
            tmp.next = list2;
        }
        if (list2 == null) {
            tmp.next = list1;
        }

        return dummy.next;
    }
}

法2：代码不够简洁

class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }
        ListNode cur = null, head = null;
        while (list1 != null || list2 != null) {
            if (head == null) {
                if (list1.val <= list2.val) {
                    head = list1;
                    list1 = list1.next;
                } else {
                    head = list2;
                    list2 = list2.next;
                }
                cur = head;
            }
            if (list1 == null) {
                cur.next = list2;
                break;
            }
            if (list2 == null) {
                cur.next = list1;
                break;
            }
            if (list1.val <= list2.val) {
                cur.next = list1;
                list1 = list1.next;
            } else {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }

        return head;
    }
}