算法1
(链表归并排序) $O(n)$
1.先创建归并完成后的新链表虚拟头结点
2.创建新链表的尾结点
3.进行归并排序
时间复杂度
参考文献
C++ 代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
c++:
class Solution {
public:
ListNode* merge(ListNode* l1, ListNode* l2) {
//虚拟头结点,创建新的链表
auto dummy = new ListNode(-1);
//尾结点
auto tail = dummy;
while(l1 && l2){
if(l1->val < l2->val){
tail = tail -> next = l1;
l1 = l1->next;
}else{
tail = tail -> next = l2;
l2 = l2->next;
}
}
if(l1) tail -> next = l1;
if(l2) tail -> next = l2;
return dummy -> next;
}
};
算法1
(归并排序) $O(n)$
时间复杂度
参考文献
java 代码
/**
* 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 tail = dummy;
while(l1 != null && l2 != null){
if(l1.val < l2.val){
tail.next = l1;
tail = tail.next;
l1 = l1.next;
}else{
tail.next = l2;
tail = tail.next;
l2 = l2.next;
}
}
if(l1 != null) tail.next = l1;
if(l2 != null) tail.next = l2;
return dummy.next;
}
}