25. Reverse Nodes in k-Group

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.

解题思路：https://leetcode.windliang.cc/leetCode-25-Reverse-Nodes-in-k-Group.html

public ListNode reverseKGroup(ListNode head, int k) {
    if (head == null)
        return null;
    ListNode sub_head = head;
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode tail = dummy;
    ListNode toNull = head;
    while (sub_head != null) {
        int i = k;
        //找到子链表的尾部
        while (i - 1 > 0) {
            toNull = toNull.next;
            if (toNull == null) {
                return dummy.next;
            }
            i--;
        }
        ListNode temp = toNull.next;
        //将子链表断开
        toNull.next = null;
        ListNode new_sub_head = reverse(sub_head); 
        //将倒置后的链表接到 tail 后边
        tail.next = new_sub_head;
        //更新 tail 
        tail = sub_head; //sub_head 由于倒置其实是新链表的尾部
        sub_head = temp;
        toNull = sub_head;
        //将后边断开的链表接回来
        tail.next = sub_head;
    }
    return dummy.next;
}
public ListNode reverse(ListNode head) {
    ListNode current_head = null;
    while (head != null) {
        ListNode next = head.next;
        head.next = current_head;
        current_head = head;
        head = next;
    }
    return current_head;
}

很好很强大，我很饿

为了将头结点也一般化，我们创建一个 dummy 结点，然后整个过程主要运用三个指针， tail 指针表示已经倒置后的链表的尾部，subhead 指针表示要进行倒置的子链表，toNull 指针为了将子链表从原来链表中取下来。

一个 while 循环，让 toNull 指针走 k - 1 步使其指向子链表的尾部。中间的 if 语句就是判断当前节点数够不够 k 个了，不够的话直接返回结果就可以了。

将子链表指向 null ，脱离出来。并且用 temp 保存下一个结点的位置。

然后调用倒置函数，将子链表倒置。

接下来四步分别是，新链表接到 tail（注意下边的图 tail 是更新后的位置，之前 tail 在 dummy 的位置） 的后边；更新 tail 到新链表的尾部，也就是之前的 subhead （下图 subhead 也是更新后的位置，之前的位置参见上边的图）；sub_head 更新到 temp 的位置；toNull 到 sub_head 的位置；然后将新的尾部 tail 把之前断开的链表连起来，接到 sub_head 上。

整理下其实就是下边的样子

和初始的时候（下边的图）对比一下，发现 tail，subhead 和 toNull 三个指针已经就位，可以愉快的重复上边的步骤了。

 

public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (k <= 1) return head;
        int T = GetLength(head) / k;
        ListNode dummy = new ListNode(0), cur = head, ins = dummy; 
        dummy.next = head;
        while ((T--) != 0) {
            for (int i = 0; i < k - 1; ++i) {
                ListNode move = cur.next;
                cur.next = move.next;
                move.next = ins.next;
                ins.next = move;
            }
            ins = cur;
            cur = cur.next;
        }
        return dummy.next;
    }
    
    public int GetLength(ListNode head) {
        int length = 0;
        while (head != null) {
            head = head.next;
            length++;
        }
        return length;
    }
}

Much better、

 

 

 

 

 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (k <= 1) return head;
        int length = 0;
        ListNode d = head;
        while(d != null){
            length++;
            d = d.next;
        }
        int T = length / k;
        ListNode dummy = new ListNode(0), cur = head, prev = dummy; 
        dummy.next = head;
        while ((T--) != 0) {
            for (int i = 0; i < k - 1; ++i) {
                ListNode move = cur.next;
                cur.next = move.next;
                move.next = prev.next;
                prev.next = move;
            }
            prev = cur;
            cur = cur.next;
        }
        return dummy.next;
    }
}