LeetCode 234. 回文链表 Palindrome Linked List (Easy)

/**
 * 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) {

        // 1.快慢指针找到中点
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast != nullptr && fast->next != nullptr)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        // 偶数节点, 慢指针走到中点
        if (fast != nullptr) slow = slow->next;

        // 2.反转后半部分链表
        ListNode* prev = nullptr;
        while (slow != nullptr)
        {
            ListNode* temp = slow->next;
            slow->next = prev;
            prev = slow;
            slow = temp;
        }

        // 3.判断两个链表是否相等
        while (head != nullptr && prev != nullptr)
        {
            if (head->val != prev->val) return false;
            head = head->next;
            prev = prev->next;
        }
        return true;
    }
};