跳转至

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？

解法

方法一：模拟

我们可以根据题意，模拟整个翻转的过程。

首先，我们定义一个辅助函数 $\textit{reverse}$，用于翻转一个链表。然后，我们定义一个虚拟头结点 $\textit{dummy}$，并将其 $\textit{next}$ 指针指向 $\textit{head}$。

接着，我们遍历链表，每次遍历 $k$ 个节点，若剩余节点不足 $k$ 个，则不进行翻转。否则，我们将 $k$ 个节点取出，然后调用 $\textit{reverse}$ 函数翻转这 $k$ 个节点。然后将翻转后的链表与原链表连接起来。继续遍历下一个 $k$ 个节点，直到遍历完整个链表。

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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        def reverse(head: Optional[ListNode]) -> Optional[ListNode]:
            dummy = ListNode()
            cur = head
            while cur:
                nxt = cur.next
                cur.next = dummy.next
                dummy.next = cur
                cur = nxt
            return dummy.next

        dummy = pre = ListNode(next=head)
        while pre:
            cur = pre
            for _ in range(k):
                cur = cur.next
                if cur is None:
                    return dummy.next
            node = pre.next
            nxt = cur.next
            cur.next = None
            pre.next = reverse(node)
            node.next = nxt
            pre = node
        return dummy.next

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0, head);
        dummy.next = head;
        ListNode pre = dummy;
        while (pre != null) {
            ListNode cur = pre;
            for (int i = 0; i < k; i++) {
                cur = cur.next;
                if (cur == null) {
                    return dummy.next;
                }
            }
            ListNode node = pre.next;
            ListNode nxt = cur.next;
            cur.next = null;
            pre.next = reverse(node);
            node.next = nxt;
            pre = node;
        }
        return dummy.next;
    }

    private ListNode reverse(ListNode head) {
        ListNode dummy = new ListNode();
        ListNode cur = head;
        while (cur != null) {
            ListNode nxt = cur.next;
            cur.next = dummy.next;
            dummy.next = cur;
            cur = nxt;
        }
        return dummy.next;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseKGroup(head *ListNode, k int) *ListNode {
    dummy := &ListNode{Next: head}
    pre := dummy

    for pre != nil {
        cur := pre
        for i := 0; i < k; i++ {
            cur = cur.Next
            if cur == nil {
                return dummy.Next
            }
        }

        node := pre.Next
        nxt := cur.Next
        cur.Next = nil
        pre.Next = reverse(node)
        node.Next = nxt
        pre = node
    }
    return dummy.Next
}

func reverse(head *ListNode) *ListNode {
    var dummy *ListNode
    cur := head
    for cur != nil {
        nxt := cur.Next
        cur.Next = dummy
        dummy = cur
        cur = nxt
    }
    return dummy
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function reverseKGroup(head: ListNode | null, k: number): ListNode | null {
    const dummy = new ListNode(0, head);
    let pre = dummy;
    while (pre !== null) {
        let cur: ListNode | null = pre;
        for (let i = 0; i < k; i++) {
            cur = cur?.next || null;
            if (cur === null) {
                return dummy.next;
            }
        }

        const node = pre.next;
        const nxt = cur?.next || null;
        cur!.next = null;
        pre.next = reverse(node);
        node!.next = nxt;
        pre = node!;
    }

    return dummy.next;
}

function reverse(head: ListNode | null): ListNode | null {
    let dummy: ListNode | null = null;
    let cur = head;

    while (cur !== null) {
        const nxt = cur.next;
        cur.next = dummy;
        dummy = cur;
        cur = nxt;
    }

    return dummy;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn reverse_k_group(head: Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {
        fn reverse(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
            let mut head = head;
            let mut pre = None;
            while let Some(mut node) = head {
                head = node.next.take();
                node.next = pre.take();
                pre = Some(node);
            }
            pre
        }

        let mut dummy = Some(Box::new(ListNode::new(0)));
        let mut pre = &mut dummy;
        let mut cur = head;
        while cur.is_some() {
            let mut q = &mut cur;
            for _ in 0..k - 1 {
                if q.is_none() {
                    break;
                }
                q = &mut q.as_mut().unwrap().next;
            }
            if q.is_none() {
                pre.as_mut().unwrap().next = cur;
                return dummy.unwrap().next;
            }

            let b = q.as_mut().unwrap().next.take();
            pre.as_mut().unwrap().next = reverse(cur);
            while pre.is_some() && pre.as_mut().unwrap().next.is_some() {
                pre = &mut pre.as_mut().unwrap().next;
            }
            cur = b;
        }
        dummy.unwrap().next
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val = 0, ListNode next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode ReverseKGroup(ListNode head, int k) {
        var dummy = new ListNode(0);
        dummy.next = head;
        var pre = dummy;

        while (pre != null) {
            var cur = pre;
            for (int i = 0; i < k; i++) {
                if (cur.next == null) {
                    return dummy.next;
                }
                cur = cur.next;
            }

            var node = pre.next;
            var nxt = cur.next;
            cur.next = null;
            pre.next = Reverse(node);
            node.next = nxt;
            pre = node;
        }

        return dummy.next;
    }

    private ListNode Reverse(ListNode head) {
        ListNode prev = null;
        var cur = head;
        while (cur != null) {
            var nxt = cur.next;
            cur.next = prev;
            prev = cur;
            cur = nxt;
        }
        return prev;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val = 0, $next = null) {
 *         $this->val = $val;
 *         $this->next = $next;
 *     }
 * }
 */
class Solution {
    /**
     * @param ListNode $head
     * @param Integer $k
     * @return ListNode
     */
    function reverseKGroup($head, $k) {
        $dummy = new ListNode(0);
        $dummy->next = $head;
        $pre = $dummy;

        while ($pre !== null) {
            $cur = $pre;
            for ($i = 0; $i < $k; $i++) {
                if ($cur->next === null) {
                    return $dummy->next;
                }
                $cur = $cur->next;
            }

            $node = $pre->next;
            $nxt = $cur->next;
            $cur->next = null;
            $pre->next = $this->reverse($node);
            $node->next = $nxt;
            $pre = $node;
        }

        return $dummy->next;
    }

    /**
     * Helper function to reverse a linked list.
     * @param ListNode $head
     * @return ListNode
     */
    function reverse($head) {
        $prev = null;
        $cur = $head;
        while ($cur !== null) {
            $nxt = $cur->next;
            $cur->next = $prev;
            $prev = $cur;
            $cur = $nxt;
        }
        return $prev;
    }
}

评论