版权声明: https://blog.csdn.net/ysq96/article/details/89918244
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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
题意:将两个有序的链表合并成为一个有序的链表
C++:
/**
* 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) {
ListNode* head=new ListNode(-1),*cur=head;
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 head->next;
}
};
Python3:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
res=ListNode(None)
node=res
while l1 and l2:
if l1.val<l2.val:
node.next,l1=l1,l1.next
else:
node.next,l2=l2,l2.next
node=node.next
if l1:
node.next=l1;
else:
node.next=l2;
return res.next