【LeetCode】Reverse Nodes in k-Group

题目

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

解答

题目要求对链表的每K个元素，翻转一次。最后一次如果不到K，则不用翻转。

可以通过链表的区间内翻转实现，遇到k组的节点就把前k个节点翻转，代码如下：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public static ListNode reverseKGroup(ListNode head,int k){    if(head==null||k==1){    return head;    }    ListNode fakeHead=new ListNode(-1);    fakeHead.next=head;    ListNode pre=fakeHead;    int i=0;    while(head!=null){    i++;    if(i%k==0){    pre=reverse(pre,head.next);  //reverse返回的last就是pre    head=pre.next;    }else{    head=head.next;    }    }    return fakeHead.next;    }    private static ListNode reverse(ListNode pre,ListNode next){  //区间的链表反转    ListNode last=pre.next;  //last不变    ListNode cur=last.next;  //cur移动    while(cur!=next){    last.next=cur.next;    cur.next=pre.next;    pre.next=cur;    cur=last.next;    }    return last;    }}

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