206.reverse-linked-list.md

题目地址

https://leetcode.com/problems/reverse-linked-list/description/

题目描述

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

思路

这个就是常规操作了，使用一个变量记录前驱 pre，一个变量记录后继 next.

不断更新current.next = pre 就好了

关键点解析

• 链表的基本操作（交换）
• 虚拟节点 dummy 简化操作
• 注意更新 current 和 pre 的位置， 否则有可能出现溢出

代码

语言支持：JS, C++, Python,Java

JavaScript Code：

/**
 * 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;

    let cur = head;
    let pre = null;

    while(cur) {
        const next = cur.next;
        cur.next = pre;
        pre = cur;
        cur = next;
    }

    return pre;
};

C++ Code：

/**
 * 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* prev = NULL;
        ListNode* cur = head;
        ListNode* next = NULL;
        while (cur != NULL) {
            next = cur->next;
            cur->next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
};

Python Code:

# 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 not head: return None
        prev = None
        cur = head
        while cur:
            cur.next, prev, cur = prev, cur, cur.next
        return prev

Java Code:

/**
 * 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 pre = null, cur = head;

        while (cur != null) {
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }

        return pre;
    }
}

拓展

通过单链表的定义可以得知，单链表也是递归结构，因此，也可以使用递归的方式来进行 reverse 操作。

由于单链表是线性的，使用递归方式将导致栈的使用也是线性的，当链表长度达到一定程度时，递归会导致爆栈，因此，现实中并不推荐使用递归方式来操作链表。

描述

1. 除第一个节点外，递归将链表 reverse
2. 将第一个节点添加到已 reverse 的链表之后

这里需要注意的是，每次需要保存已经 reverse 的链表的头节点和尾节点

C++实现

// 普通递归
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* tail = nullptr;
        return reverseRecursive(head, tail);
    }

    ListNode* reverseRecursive(ListNode *head, ListNode *&tail) {
        if (head == nullptr) {
            tail = nullptr;
            return head;
        }
        if (head->next == nullptr) {
            tail = head;
            return head;
        }
        auto h = reverseRecursive(head->next, tail);
        if (tail != nullptr) {
            tail->next = head;
            tail = head;
            head->next = nullptr;
        }
        return h;
    }
};

// （类似）尾递归
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == nullptr) return head;
        return reverseRecursive(nullptr, head, head->next);
    }

    ListNode* reverseRecursive(ListNode *prev, ListNode *head, ListNode *next)
    {
        if (next == nullptr) return head;
        auto n = next->next;
        next->next = head;
        head->next = prev;
        return reverseRecursive(head, next, n);
    }
};

JavaScript 实现

var reverseList = function(head) {
  // 递归结束条件
  if (head === null || head.next === null) {
    return head
  }

  // 递归反转 子链表
  let newReverseList = reverseList(head.next)
  // 获取原来链表的第 2 个节点 newReverseListTail
  let newReverseListTail = head.next
  // 调整原来头结点和第 2 个节点的指向
  newReverseListTail.next = head
  head.next = null

  // 将调整后的链表返回
  return newReverseList
}

Python 实现

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if not head or not head.next: return head
        ans = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return ans