LeetCode 206. Reverse Linked List 题解（C++）_leetcode 简略题解c语言

LeetCode 206. Reverse Linked List 题解（C++）

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题目描述

• Reverse a singly linked list.

思路

迭代

• 设置三个指针分别指向当前结点，当前结点的前一结点和当前结点的后一结点，之后遍历整个链表，使当前结点指向上一结点，并依次将三个指针向下一结点移动。

递归

• 对当前结点的下一结点做递归，返回倒置后的链表，之后使当前结点的下一结点指向当前结点，当前结点指向NULL。

代码

迭代

class Solution
{
public:
    ListNode* reverseList(ListNode* head) 
    {
        if (head == NULL)
        {
            return NULL;
        }
        ListNode *previousNode = NULL;
        ListNode *currentNode = head;
        ListNode *nextNode = head->next;
        while (nextNode != NULL)
        {
            currentNode->next = previousNode;
            previousNode = currentNode;
            currentNode = nextNode;
            nextNode = currentNode->next;
        }
        currentNode->next = previousNode;
        return currentNode;
    }
};
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递归

class Solution
{
public:
    ListNode* reverseList(ListNode* head) 
    {
        if (head == NULL || head->next == NULL)
        {
            return head;
        }
        ListNode *p = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;
        return p;
    }
};
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