25. Reverse Nodes in k-Group

题目

原始地址：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        
    }
}

描述

给定一个单链表，每k个节点做一次反转，并返回反转过的链表。

分析

基本思路就是每k个节点拆分出一个子链表并且做反转，需要注意反转后子链表的前后要和其它部分正确衔接。给出循环和递归两种解法。

解法1

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0), curr = head, prev = dummy, start, end;
        dummy.next = head;
        int i = 0;
        while (curr != null) {
            if (++i % k == 0) {
                start = prev.next;
                end = curr.next;
                curr.next = null;
                prev.next = reverse(start);
                start.next = end;
                prev = start;
                curr = end;
            } else {
                curr = curr.next;
            }
        }
        return dummy.next;
    }

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

解法2

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode curr = head, prev = null;
        int i = 0;
        while (i < k && curr != null) {
            i++;
            prev = curr;
            curr = curr.next;
        }
        if (i < k) {
            return head;
        } else {
            prev.next = null;
            reverse(head);
            head.next = reverseKGroup(curr, k);
            return prev;
        }
    }

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