以K个为一组反转单链表,最后不足K个节点的部分也反转

package StackMin.ReverseList_offer16;

public class ReverseKgroup_extend_offer16 {
    /**
     * 分组反转单链表,最后不足K个节点的部分也反转
     *
     * @param head
     * @param k
     * @return
     */
    public static ListNode reverseKgroup(ListNode head, int k) {
        if (head == null)
            return head;
        ListNode cur = head;
        ListNode reHead = null;

        int count = 0;
        /* Reverse first k nodes of linked list */
        while (count < k && cur != null) {
            ListNode reCur = cur;
            cur = cur.next;
            reCur.next = reHead;
            reHead = reCur;
            count++;
        }
        /*
         * cur is now a pointer to (k+1)th node Recursively call for the
         * list starting from current. And make rest of the list as next of
         * first node
         */
        if (cur != null)
            head.next = reverseKgroup(cur, k);

        return reHead;
    }

}
package StackMin.ReverseList_offer16;

public class ReverseKgroup_extend_offer16 {
    /**
     * 分组反转单链表,最后不足K个节点的部分也反转
     *
     * @param head
     * @param k
     * @return
     */
    /*
    输入的原始单链表为3-5-6-9-7-2-1-12,其中K为3;
    经过第一次while循环,单链表变为6-5-3-9-7-2-1-12。此时跳出while循环是因为count<k不成立了,cur节点指向了9,head节点指向了3。
    所以接着判断cur是否为null,若不是,则刚好递归求出head.next。
    经过第二次while循环,单链表为6-5-3-2-7-9-1-12。此时跳出while循环是因为count<k不成立了,cur节点指向了1,head节点指向了9。
    接着判断cur,并且递归求head.next节点。
    第三次循环,跳出while是因为cur==null了,直接返回reHead,此时reHead指向了12。
     */
    public ListNode reverseKgroup(ListNode head, int k) {
        if (head == null)
            return head;
        ListNode cur = head;
        ListNode reHead = null;

        int count = 0;
        /* Reverse first k nodes of linked list */
        while (count < k && cur != null) {
            ListNode reCur = cur;
            cur = cur.next;
            reCur.next = reHead;
            reHead = reCur;
            count++;
        }
        /*
         * cur is now a pointer to (k+1)th node Recursively call for the
         * list starting from current. And make rest of the list as next of
         * first node
         */
        if (cur != null)
            head.next = reverseKgroup(cur, k);

        return reHead;
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(3);
        ListNode L2 = new ListNode(5);
        ListNode L3 = new ListNode(6);
        ListNode L4 = new ListNode(9);
        ListNode L5 = new ListNode(7);
        ListNode L6 = new ListNode(2);
        ListNode L7 = new ListNode(1);
        ListNode L8 = new ListNode(12);
        head.next = L2;
        L2.next = L3;
        L3.next = L4;
        L4.next = L5;
        L5.next = L6;
        L6.next = L7;
        L7.next = L8;
        ReverseKgroup_extend_offer16 reverseKgroup_extend_offer16 = new ReverseKgroup_extend_offer16();
        reverseKgroup_extend_offer16.printListNode(head);
        System.out.println();
        ListNode node = reverseKgroup_extend_offer16.reverseKgroup(head, 3);
        reverseKgroup_extend_offer16.printListNode(node);


    }

    public void printListNode(ListNode node) {
        while (node != null) {
            if (node.next == null) {
                System.out.print(node.val);

            } else {
                System.out.print(node.val + ">");
            }
            node = node.next;
        }
    }

}
package StackMin.ReverseList_offer16;

public class ListNode {
    int val;
    public ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }

}
原文地址:https://www.cnblogs.com/chengpeng15/p/9872670.html