21. 合并两个有序链表

21. 合并两个有序链表 #

链接 #

题目 #

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]

提示：

两个链表的节点数目范围是 [0, 50]
-100 <= Node.val <= 100
l1 和 l2 均按 非递减顺序 排列

解答 #

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode dummy;
        ListNode* cur = &dummy;
        while(l1 && 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 ? l1 : l2;

        return dummy.next;
    }
};

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]):
        d = ListNode()
        c = d
        while l1 and l2:
            if l1.val < l2.val:
                c.next = l1
                l1 = l1.next
            else:
                c.next = l2
                l2 = l2.next

            c = c.next

        c.next = l1 if l1 else l2

        return d.next