25. Reverse Nodes in k-Group (JAVA)
Given a linked list, reverse the nodes of a linked list k at a time and return its 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 kthen left-out nodes in the end should remain as it is.
Example:
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
Note:
- Only constant extra memory is allowed.
- You may not alter the values in the list's nodes, only nodes itself may be changed.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode reverseKGroup(ListNode head, int k) { if(k == 1) return head; ListNode cur = head; ListNode curHead = head; //head of current subgroup ListNode nextHead = head; //head of next subgroup ListNode preTail = null; //tail of previous subgroup while(true){ //check whether there's enough elements in subgroup for(int i = 1; i < k; i++){ if(cur == null) break; //not enough element in subgroup cur = cur.next; } if(cur == null) break; //not enough element in subgroup //reverse nodes cur = curHead.next; for(int i = 1; i < k; i++){ nextHead = cur.next; if(preTail==null) { cur.next = head; head = cur; } else { cur.next = preTail.next; preTail.next = cur; } cur = nextHead; } curHead.next = nextHead; preTail = curHead; curHead = nextHead; } return head; } }
因为while语句是无条件循环,特别要注意k==1时的死循环。