Given a linked list, reverse the nodes of a linked list k at a time and return its modified 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 nodes, only nodes itself may be changed.
Only constant memory is allowed.
For 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
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode reverseKGroup(ListNode head, int k) { if(head == null || k == 1){ return head; } int len = 0; ListNode p = head; while(p != null){ p = p.next; len++; } ListNode safeG = new ListNode(-1); safeG.next = head; ListNode pre = safeG, cur = head, post = head.next; int m = len/k; for(int i = 0; i < m; i++){ post = cur.next; int j = 0; while(post != null){ ListNode tmp = post.next; post.next = cur; cur = post; post = tmp; j++; if(j == k-1){ break; } } ListNode tmp = pre.next; pre.next =cur; tmp.next = post; pre = tmp; cur = pre.next; } return safeG.next; } }
[Leetcode]-- Reverse Nodes in k-Group
原文：http://www.cnblogs.com/RazerLu/p/3538143.html