Question

206. 反转链表

English Version

题目描述

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

 

示例 1：

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

示例 2：

输入：head = [1,2]
输出：[2,1]

示例 3：

输入：head = []
输出：[]

 

提示：

• 链表中节点的数目范围是 [0, 5000]
• -5000 <= Node.val <= 5000

 

进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？

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, val=0, next=None):
#     self.val = val
#     self.next = next
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() {}
 *   ListNode(int val) { this.val = val; }
 *   ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
  public ListNode reverseList(ListNode head) {
    ListNode dummy = new ListNode();
    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() : val(0), next(nullptr) {}
 *   ListNode(int x) : val(x), next(nullptr) {}
 *   ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* reverseList(ListNode* head) {
    ListNode* dummy = new ListNode();
    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 {
  if (head == null) {
    return head;
  }
  let pre = null;
  let cur = head;
  while (cur != null) {
    const next = cur.next;
    cur.next = pre;
    [pre, cur] = [cur, next];
  }
  return pre;
}

// 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 head = head;
    let mut pre = None;
    while let Some(mut node) = head {
      head = node.next.take();
      node.next = pre.take();
      pre = Some(node);
    }
    pre
  }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
  let dummy = new ListNode();
  let curr = head;
  while (curr) {
    let 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 val=0, ListNode next=null) {
 *     this.val = val;
 *     this.next = next;
 *   }
 * }
 */
public class Solution {
  public ListNode ReverseList(ListNode head) {
    ListNode pre = null;
    for (ListNode p = head; p != null;)
    {
      ListNode t = p.next;
      p.next = pre;
      pre = p;
      p = t;
    }
    return pre;
  }
}

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, val=0, next=None):
#     self.val = val
#     self.next = next
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() {}
 *   ListNode(int val) { this.val = val; }
 *   ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
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() : val(0), next(nullptr) {}
 *   ListNode(int x) : val(x), next(nullptr) {}
 *   ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
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)
 *   }
 * }
 */
const rev = (pre: ListNode | null, cur: ListNode | null): ListNode | null => {
  if (cur == null) {
    return pre;
  }
  const next = cur.next;
  cur.next = pre;
  return rev(cur, next);
};

function reverseList(head: ListNode | null): ListNode | null {
  if (head == null) {
    return head;
  }
  const next = head.next;
  head.next = null;
  return rev(head, 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 {
  fn rev(pre: Option<Box<ListNode>>, cur: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    match cur {
      None => pre,
      Some(mut node) => {
        let next = node.next;
        node.next = pre;
        Self::rev(Some(node), next)
      }
    }
  }

  pub fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    Self::rev(None, head)
  }
}