25. K 个一组翻转链表

题目描述

给你链表的头节点 head ，每 k 个节点一组进行翻转，请你返回修改后的链表。

k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。

你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。

 

示例 1：

输入：head = [1,2,3,4,5], k = 2
输出：[2,1,4,3,5]

示例 2：

输入：head = [1,2,3,4,5], k = 3
输出：[3,2,1,4,5]

 

提示：

• 链表中的节点数目为 n
• 1 <= k <= n <= 5000
• 0 <= Node.val <= 1000

 

进阶：你可以设计一个只用 O(1) 额外内存空间的算法解决此问题吗？

方法一：迭代

时间复杂度为 $O(n)$，空间复杂度为 $O(1)$，其中 $n$ 是链表的长度。

Java

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0, head);
        ListNode pre = dummy, cur = dummy;
        while (cur.next != null) {
            for (int i = 0; i < k && cur != null; ++i) {
                cur = cur.next;
            }
            if (cur == null) {
                return dummy.next;
            }
            ListNode t = cur.next;
            cur.next = null;
            ListNode start = pre.next;
            pre.next = reverseList(start);
            start.next = t;
            pre = start;
            cur = pre;
        }
        return dummy.next;
    }

    private ListNode reverseList(ListNode head) {
        ListNode pre = null, p = head;
        while (p != null) {
            ListNode q = p.next;
            p.next = pre;
            pre = p;
            p = q;
        }
        return pre;
    }
}

TypeScript

function reverseKGroup(head: ListNode | null, k: number): ListNode | null {
    let dummy = new ListNode(0, head);
    let pre = dummy;
    // pre->head-> ... ->tail-> next
    while (head != null) {
        let tail = pre;
        for (let i = 0; i < k; ++i) {
            tail = tail.next;
            if (tail == null) {
                return dummy.next;
            }
        }
        let t = tail.next;
        [head, tail] = reverse(head, tail);
        // set next
        pre.next = head;
        tail.next = t;
        // set new pre and new head
        pre = tail;
        head = t;
    }
    return dummy.next;
}

function reverse(head: ListNode, tail: ListNode) {
    let cur = head;
    let pre = tail.next;
    // head -> next -> ... -> tail -> pre
    while (pre != tail) {
        let t = cur.next;
        cur.next = pre;
        pre = cur;
        cur = t;
    }
    return [tail, head];
}

Python

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        def reverseList(head):
            pre, p = None, head
            while p:
                q = p.next
                p.next = pre
                pre = p
                p = q
            return pre

        dummy = ListNode(next=head)
        pre = cur = dummy
        while cur.next:
            for _ in range(k):
                cur = cur.next
                if cur is None:
                    return dummy.next
            t = cur.next
            cur.next = None
            start = pre.next
            pre.next = reverseList(start)
            start.next = t
            pre = start
            cur = pre
        return dummy.next

方法二：递归

时间复杂度为 $O(n)$，空间复杂度为 $O({\log }_{k}n)$，其中 $n$ 是链表的长度。

TypeScript

function reverseKGroup(head: ListNode | null, k: number): ListNode | null {
    if (k === 1) {
        return head;
    }

    const dummy = new ListNode(0, head);
    let root = dummy;
    while (root != null) {
        let pre = root;
        let cur = root;

        let count = 0;
        while (count !== k) {
            count++;
            cur = cur.next;
            if (cur == null) {
                return dummy.next;
            }
        }

        const nextRoot = pre.next;
        pre.next = cur;

        let node = nextRoot;
        let next = node.next;
        node.next = cur.next;
        while (node != cur) {
            [next.next, node, next] = [node, next, next.next];
        }
        root = nextRoot;
    }

    return dummy.next;
}