Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.
You may not alter the values in the list's nodes, only nodes themselves may be changed.
Example 1:
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]
Example 2:
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]
Constraints:
- The number of nodes in the list is n.
- 1 <= k <= n <= 5000
- 0 <= Node.val <= 1000
Follow-up: Can you solve the problem in O(1) extra memory space?
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if(head==null || head.next==null){
return head;
}
int count = 0;
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy;
ListNode cur = head;
while(cur!=null){
ListNode next = cur.next;
count++;
if(count==k){
pre = reverse(pre, next);
count=0;
}
cur = next;
}
return dummy.next;
}
private ListNode reverse(ListNode pre, ListNode end){
if(pre==null || pre.next==null){
return pre;
}
ListNode head = pre.next;
ListNode cur = pre.next.next;
while(cur != next){
ListNode next = cur.next;
cur.next = pre.next;
pre.next = cur;
cur = next;
}
//end其实是下一个要反转的头结点
head.next = end;
return head;
}
}