21. Merge Two Sorted Lists

最新推荐文章于 2025-09-10 15:24:29 发布

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最新推荐文章于 2025-09-10 15:24:29 发布

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文章标签：数据结构与算法 java python

本文详细介绍了如何使用合并两个排序的单链表的算法，包括使用虚拟头节点的方法来实现链表合并，并提供了Java和Python的实现代码。通过实例分析，解释了时间复杂度为O(n)和空间复杂度为O(1)的合并过程。

题目：

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

链接： http://leetcode.com/problems/merge-two-sorted-lists/

题解：合并两个排序的单链表，创建一个dummy head，然后进行合并。 Time Complexity - O(n)，Space Complexity - O(1)。

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null)
            return l2;
        if(l2 == null)
            return l1;
        ListNode result = new ListNode(-1);
        ListNode node = result;
        
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                node.next = l1;
                l1 = l1.next;
            }
            else{
                node.next = l2;
                l2 = l2.next;
            }
            node = node.next;
            node.next = null;
        }
        
        if(l1 != null)
            node.next = l1;
        else if (l2 != null)
            node.next = l2;
        return result.next;
    }
}

二刷：

Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode node = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                node.next = l1;
                l1 = l1.next;
            } else {
                node.next = l2;
                l2 = l2.next;
            }
            node = node.next;
        }
        node.next = l1 != null ? l1 : l2;
        return dummy.next;
    }
}

Python:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(-1)
        node = dummy
        while l1 != None and l2 != None:
            if l1.val < l2.val:
                node.next = l1
                l1 = l1.next
            else:
                node.next = l2
                l2 = l2.next
            node = node.next
        node.next = l1 if l1 != None else l2
        return dummy.next

三刷:

Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode node = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                node.next = l1;
                l1 = l1.next;
            } else {
                node.next = l2;
                l2 = l2.next;
            }
            node = node.next;
        }
        node.next = (l1 != null) ? l1 : l2;
        return dummy.next;
    }
}