AcWing 36. 合并两个排序的链表
原题链接
简单
作者:
STU756
,
2021-01-20 17:55:06
,
所有人可见
,
阅读 357
//java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode merge(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode cur = dummy;
while(l1!=null && l2!=null) {
if(l1.val <= l2.val) {
cur.next = l1;
l1 = l1.next;
}else {
cur.next = l2;
l2 = l2.next;
}
cur = cur.next;
}
cur.next = (l1== null? l2 : l1);
return dummy.next;
}
}
//c++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* merge(ListNode* l1, ListNode* l2) {
ListNode * dummy = new ListNode(-1);
ListNode *cur = dummy;
while(l1!=NULL && l2 != NULL) {
if(l1->val <= l2->val) {
cur -> next = l1;
l1 = l1->next;
}else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
cur ->next = (l1!= NULL? l1 : l2);
return dummy->next;
}
};