21. Merge Two Sorted Lists

最新推荐文章于 2024-01-12 14:30:52 发布

chuandalishuo

最新推荐文章于 2024-01-12 14:30:52 发布

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分类专栏： Leetcode 链表文章标签： Leetcode 链表

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本文提供了一种将两个已排序的链表合并为一个新排序链表的方法，并给出了Java及Python实现代码。新链表通过拼接两个输入链表的节点生成。

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.

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode p=head, p1=l1, p2=l2;
        while(p1!=null&&p2!=null){
            if(p1.val<p2.val){p.next=p1; p1=p1.next;}
            else{p.next=p2; p2=p2.next;}
            p=p.next;
        }
        if(p1!=null) p.next = p1;
        else if(p2!=null) p.next=p2;
        return head.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
        """
        head=ListNode(0); dup=head;
        while l1!=None and l2!=None :
            if l1.val < l2.val :
                dup.next=l1
                l1 = l1.next
                dup = dup.next
            else:
                dup.next=l2
                l2 = l2.next
                dup = dup.next
        
        if l2==None and l1!=None:
            dup.next=l1
        elif l1==None and l2!=None:
            dup.next=l2
        return head.next