回文链表

题目链接：回文链表

给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。

方法一：额外数组

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        vector<int> vals;
        while (head != nullptr) {
            vals.emplace_back(head->val);
            head = head->next;
        }
        for (int i = 0, j = (int)vals.size() - 1; i < j; ++i, --j) {
            if (vals[i] != vals[j]) {
                return false;
            }
        }
        return true;
    }
};

方法二：反转链表+找中间节点

/**
 * 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) {}
 * };
 */
class Solution {
    ListNode* findmid(ListNode* head) {
        ListNode* slow = head,*fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
    ListNode* reserve(ListNode* head) {
        ListNode* ans = nullptr;
        ListNode* cur = head;
        while(cur) {
            ListNode* t = cur->next;
            cur->next = ans;
            ans = cur;
            cur = t;
        }
        return ans;
    }
public:
    bool isPalindrome(ListNode* head) {
        ListNode* head1 = findmid(head);
        ListNode* head2 = reserve(head1);
        while (head2) {
            if (head->val != head2->val) {
                return false;
            }
            head = head->next;
            head2 = head2->next;
        }
        return true;
    }
};