欢迎使用优快云-markdown编辑器

最新推荐文章于 2021-01-07 23:01:28 发布

原创最新推荐文章于 2021-01-07 23:01:28 发布 · 282 阅读

0 ·

CC 4.0 BY-SA版权

java 同时被 2 个专栏收录

6 篇文章

订阅专栏

leetcode

6 篇文章

订阅专栏

本文介绍了一种算法，用于合并两个已排序的链表并返回一个新的有序链表。该算法通过遍历两个输入链表，并比较节点值来构建新的链表。

21. Merge Two Sorted Lists

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.

代码块

代码块语法遵循标准markdown代码，例如：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {  
public:  
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {  
        if (l1 == NULL)  
            return l2;  
        if (l2 == NULL)  
            return l1;  

        ListNode *start, *cur;  

        if (l1->val < l2->val) {  
            cur = start = l1;  
            l1 = l1->next;  
        } else {  
            cur = start = l2;  
            l2 = l2->next;  
        }  
        while (l1 != NULL && l2 != NULL) {  
            if (l1->val < l2->val) {  
                cur->next = l1;  
                cur = l1;  
                l1 = l1->next;  
            } else {  
                cur->next = l2;  
                cur = l2;  
                l2 = l2->next;  
            }  
        }  
        if (l1 != NULL)  
            cur->next = l1;  
        else  
            cur->next = l2;  
        return start;  
    }  
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {  
public:  
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l2 == NULL) return l1;
        if(l1 == NULL) return l2;

        if(l1->val > l2->val) {
            ListNode* temp = l2;
            temp->next = mergeTwoLists(l1, l2->next);
            return temp;
        } else {
            ListNode* temp = l1;
            temp->next = mergeTwoLists(l1->next, l2);
            return temp;
        }
    }
};

# 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
        """
        if not l1 and not l2:  return None
        dummy = ListNode(0)
        cur = dummy
        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 or l2
        return dummy.next