LeetCode题解206.reverse-linked-list

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

JavaScript Code：

/*
 * @lc app=leetcode id=206 lang=javascript
 *
 * [206] Reverse Linked List
 *
 * https://leetcode.com/problems/reverse-linked-list/description/
 *
 * algorithms
 * Easy (52.95%)
 * Total Accepted:    532.6K
 * Total Submissions: 1M
 * Testcase Example:  '[1,2,3,4,5]'
 *
 * 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?
 *
 */
/**
 * 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

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

除第一个节点外，递归将链表reverse
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
}

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