LeetCode-206. Reverse Linked List [C++][Java]

LeetCode-206. Reverse Linked ListLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/reverse-linked-list/

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

Input: head = [1,2]
Output: [2,1]

Example 3:

Input: head = []
Output: []

Constraints:

• The number of nodes in the list is the range [0, 5000].
• -5000 <= Node.val <= 5000

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

【C++】

/**
 * 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) {}
 * };
 */

1. 递归

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

2. 循环

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *prev = nullptr, *next;
        while (head) {
            next = head->next;
            head->next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
};

【Java】

/**
 * 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; }
 * }
 */

1. 递归

class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) {return head;}
        ListNode newHead = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return newHead;
    }
}

2. 循环

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null, next;
        while (head != null) {
            next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
}

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