面试题24

定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

限制：

• 0 <= 节点个数 <= 5000

注意：本题与主站 206 题相同：https://leetcode-cn.com/problems/reverse-linked-list/

Solutions

1. insertion at front

/**
 * 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) {
        if (!head) return nullptr;
        ListNode dummy;

        while (head) {
            ListNode * next = head->next;
            head->next = dummy.next;
            dummy.next = head;
            head = next;
        }

        return dummy.next;
    }
};

1. reverse links

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (!head || !head->next) return head;

        ListNode * prev = head, * cur = head->next;
        head->next = nullptr;
        while (cur) {
            ListNode * nnext = cur->next;
            cur->next = prev;
            prev = cur;
            cur = nnext;
        }

        return prev;
    }
};

1. recursion

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (!head) return nullptr;
        if (!head->next) return head;
        ListNode * newhead = reverseList(head->next);
        head->next->next = head;
        head->next = nullptr;

        return newhead;
    }
};