leetcode系列（77）Palindrome Linked List

Given a singly linked list, determine if itis a palindrome.

解答：快慢指针split链表，需要处理链表长度为奇数（fast肯定是在最后一个节点停止）和长度为偶数（fast指针肯定是在nullptr停止）两种情况；然后reverse and compare。

class Solution {public:    bool isPalindrome(ListNode* head) {        if (head == nullptr || head->next == nullptr) {            return true;        }        // split the list        ListNode* slow = head;        ListNode* fast = head;        while (fast != nullptr && fast->next != nullptr) {            slow = slow->next;            fast = fast->next->next;        }        ListNode* ptr = (fast == nullptr ? slow : slow->next);        // reverse another half        ListNode* pre = nullptr;        auto cur = ptr;        while (cur != nullptr) {            auto post = cur->next;            cur->next = pre;            pre = cur;            cur = post;        }        ptr = pre;        // compare two half list        for (ptr = pre ; ptr != nullptr; ptr = ptr->next, head = head->next) {            if (ptr->val != head->val) {                return false;            }        }        return true;    }};