[LeetCode]Reverse Linked List

题目描述:(链接)

Given a singly linked list, determine if it is a palindrome.

解题思路：

使用快慢指针，找到链表的中心点，然后逆序后一半链表，最后再一一比较！

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if (head == nullptr) return true;
        
        ListNode *slow = head;
        ListNode *fast = head;
        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }
        
        // 逆序后一半链表
        if (fast != nullptr) {
            fast = reversedListedArray(slow->next);
        } else {
            fast = reversedListedArray(slow);
        }
        
        slow = head;
        while (slow != nullptr && fast != nullptr) {
            if (slow->val != fast->val) {
                return false;
            }
            slow = slow->next;
            fast = fast->next;
        }
        
        return true;
    }
private:
    ListNode *reversedListedArray(ListNode *head) {
        if (head == nullptr) return head;
        
        ListNode dummy(-1);
        dummy.next = head;
        ListNode *prev = dummy.next;
        ListNode *cur = prev->next;
        while (cur != nullptr) {
            prev->next = cur->next;
            cur->next = dummy.next;
            dummy.next = cur;
            cur = prev->next;
        }
        return dummy.next;
    }
};