链表操作 —— 25_k个一组翻转链表

3. 25_k个一组翻转链表

/*
给你这个链表：1->2->3->4->5

当 k = 2 时，应当返回: 2->1->4->3->5

当 k = 3 时，应当返回: 3->2->1->4->5
*/
 

/*方法一*/
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0), prev = dummy, curr = head, next;
        dummy.next = head;
        int length = 0;
        while(head != null) {
            length++;
            head = head.next;
        }
        head = dummy.next;
        for(int i = 0; i < length / k; i++) {
            for(int j = 0; j < k - 1; j++) {
                next = curr.next;
                curr.next = next.next;
                next.next = prev.next;
                prev.next = next;
            }
            prev = curr;
            curr = prev.next;
        }
        return dummy.next;
    }
}

/*方法二*/
public class Solution {

    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode hair = new ListNode(0);
        hair.next = head;

        ListNode pre = hair;
        ListNode end = hair;

        while (end.next != null) {
            for (int i = 0; i < k && end != null; i++){
                end = end.next;
            }
            if (end == null){
                break;
            }
            ListNode start = pre.next;
            ListNode next = end.next;
            end.next = null;
            pre.next = reverse(start);
            start.next = next;
            pre = start;

            end = pre;
        }
        return hair.next;
    }

    private ListNode reverse(ListNode head) {
        ListNode pre = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = pre;
            pre = curr;
            curr = next;
        }
        return pre;
    }
}

/*方法三  递归*/
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode prev = null;
        ListNode cur = head;
        ListNode next = null;
        ListNode check = head;
        int canProceed = 0;
        int count = 0;
        // 检查链表长度是否满足翻转
        while (canProceed < k && check != null) {
            check = check.next;
            canProceed++;
        }
        // 满足条件，进行翻转
        if (canProceed == k) {
            while (count < k && cur != null) {
                next = cur.next;
                cur.next = prev;
                prev = cur;
                cur = next;
                count++;
            }
            if (next != null) {
                // head 为链表翻转后的尾节点
                head.next = reverseKGroup(next, k);
            }
            // prev 为链表翻转后的头结点
            return prev;
        } else {
            // 不满住翻转条件，直接返回 head 即可
            return head;
        }
    }
}