leetcode刷题（25）——k个结点一组反转单链表

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本文介绍了一种算法，用于将链表中每K个节点进行翻转，并返回修改后的链表。此方法适用于链表长度不是K的倍数的情况，剩余节点保持原样。算法使用常数额外内存，不改变节点值，仅调整节点链接。

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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 k then 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(head==null||head.next==null||k==1){return head;}
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode cur = head;
        
        int count = 0;
        for(ListNode temp = head;temp!=null;temp = temp.next){count++;}
        
        for(int i=0;i<count/k;i++){
            for(int j=0;j<k-1;j++){
                ListNode next = cur.next;
                cur.next = next.next;
                next.next = prev.next;
                prev.next = next;
            }
            prev = cur;
            cur = prev.next;
        }
        return dummy.next;
    }
}