Question

面试题 24. 反转链表

题目描述

定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。

 

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

 

限制：

0 <= 节点个数 <= 5000

 

注意：本题与主站 206 题相同：https://leetcode.cn/problems/reverse-linked-list/

解法

方法一：头插法

创建虚拟头节点 $dummy$，遍历链表，将每个节点依次插入 $dummy$ 的下一个节点。遍历结束，返回 $dummy.next$。

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

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def reverseList(self, head: ListNode) -> ListNode:
    dummy = ListNode()
    curr = head
    while curr:
      next = curr.next
      curr.next = dummy.next
      dummy.next = curr
      curr = next
    return dummy.next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode reverseList(ListNode head) {
    ListNode dummy = new ListNode(0);
    ListNode curr = head;
    while (curr != null) {
      ListNode next = curr.next;
      curr.next = dummy.next;
      dummy.next = curr;
      curr = next;
    }
    return dummy.next;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* reverseList(ListNode* head) {
    ListNode* dummy = new ListNode(0);
    ListNode* curr = head;
    while (curr) {
      ListNode* next = curr->next;
      curr->next = dummy->next;
      dummy->next = curr;
      curr = next;
    }
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
  dummy := &ListNode{}
  curr := head
  for curr != nil {
    next := curr.Next
    curr.Next = dummy.Next
    dummy.Next = curr
    curr = next
  }
  return dummy.Next
}

/**
 * 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 reverseList(head: ListNode | null): ListNode | null {
  const dummy = new ListNode(0);
  let curr = head;
  while (curr) {
    const next = curr.next;
    curr.next = dummy.next;
    dummy.next = curr;
    curr = next;
  }
  return dummy.next;
}

// 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_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    let mut pre = None;
    let mut cur = head;

    while let Some(mut node) = cur {
      cur = node.next.take();
      node.next = pre.take();
      pre = Some(node);
    }
    pre
  }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *   this.val = val;
 *   this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
  const dummy = new ListNode(0);
  let curr = head;
  while (curr) {
    const next = curr.next;
    curr.next = dummy.next;
    dummy.next = curr;
    curr = next;
  }
  return dummy.next;
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   public int val;
 *   public ListNode next;
 *   public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
  public ListNode ReverseList(ListNode head) {
    ListNode dummy = new ListNode(0);
    ListNode curr = head;
    while (curr != null) {
      ListNode next = curr.next;
      curr.next = dummy.next;
      dummy.next = curr;
      curr = next;
    }
    return dummy.next;
  }
}

方法二：递归

递归反转链表的第二个节点到尾部的所有节点，然后 $head$ 插在反转后的链表的尾部。

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

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def reverseList(self, head: ListNode) -> ListNode:
    if head is None or head.next is None:
      return head
    ans = self.reverseList(head.next)
    head.next.next = head
    head.next = None
    return ans

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) {
      return head;
    }
    ListNode ans = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return ans;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* reverseList(ListNode* head) {
    if (!head || !head->next) {
      return head;
    }
    ListNode* ans = reverseList(head->next);
    head->next->next = head;
    head->next = nullptr;
    return ans;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
  if head == nil || head.Next == nil {
    return head
  }
  ans := reverseList(head.Next)
  head.Next.Next = head
  head.Next = nil
  return ans
}

/**
 * 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 reverseList(head: ListNode | null): ListNode | null {
  if (!head || !head.next) {
    return head;
  }
  const ans = reverseList(head.next);
  head.next.next = head;
  head.next = null;
  return ans;
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *   this.val = val;
 *   this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
  if (!head || !head.next) {
    return head;
  }
  const ans = reverseList(head.next);
  head.next.next = head;
  head.next = null;
  return ans;
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   public int val;
 *   public ListNode next;
 *   public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
  public ListNode ReverseList(ListNode head) {
    if (head == null || head.next == null) {
      return head;
    }
    ListNode ans = ReverseList(head.next);
    head.next.next = head;
    head.next = null;
    return ans;
  }
}