Algorithm--The K-order problem of linked list

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the K reverse order problem in the list List of K reverse orderThe 7th section of the list of K reverse order exercises

There is a single linked list, please design an algorithm, so that each k nodes in reverse order, if the last not enough K nodes a group, then not adjust the last few nodes. For example the list 1->2->3->4->5->6->7->8->null,k=3 this example. After adjustment for, 3->2->1->6->5->4->7->8->null. Because of k==3, the reverse is reversed between each of the three nodes, but the 7,8 is not adjusted because only two nodes are not enough.

Given the head of a single-linked list, given the K value, returns the head pointer of the linked list after the reverse order.

Java (javac 1.7)Code Auto-Complete1

Import Java. util. *;

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/*

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public class ListNode {

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    int Val;

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    ListNode next = null;

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    ListNode (int val) {

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        This.val = val;

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    }

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}*/

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 Public class Kinverse {

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     Public ListNode Inverse (listnodeheadintk) {

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        if (k<2) {

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            return head;

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        }

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        ListNode cur = head; //The current node being traversed

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        ListNode Start = null;

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        ListNode Pre = null;

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        ListNode Next = null; //Next node to traverse

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        int Count = 1;

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         while (cur!=null) {

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            Next = cur. next;

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            if (count= =k) {//count meet requirements start reverse order

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                Start = Pre == NULL ? Head pre. next;

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                Head = Pre == NULL ? cur head;

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                resign (prestartcurnext);

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                Pre = start;

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                Count = 0;

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            }

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            Count ++;

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            cur = next;

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        }

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        return head;

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    }

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     Public void resign (listnodeleftlistnodestartlistnodeEnd,

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            ListNode  Right {

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        ListNode Pre = start;

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        ListNode cur = start. next;

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        ListNode Next = null;

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         while (cur!=right) {

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            Next = cur. next;

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            cur. Next = pre;

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            Pre = cur;

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            cur = next;

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        }

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        if (left!=null) {

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            left. Next = End;

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        }

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        start. Next = right;

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    }

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}

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Algorithm--The K-order problem of linked list