leetcode 234. Palindrome Linked List

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

Follow up:
Could you do it in O(n) time and O(1) space?
判断一个链表是不是回文的。
思路:遍历节点得到总数tot,然后反转链表的后半部分,再和前半部分比较。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* head2 = nullptr;
    int tot = 0;
    int num = 0;
    void reverse(ListNode* p, ListNode* q) {
        if (p == nullptr) {
            head2 = q;
            return ;
        }
        reverse(p->next, p);
        if (q) q->next = nullptr;
        if (p) p->next = q;
    }
    bool isPalindrome(ListNode* head) {
        ListNode* p = head;
        ListNode* x = nullptr;
        ListNode* q = nullptr;
        while (p) {
            tot++;
            p = p->next;
        }
        p = head;
        while (p) {
            num++;
            if (num == tot/2) {
                x = p;
                break;
            }
            p = p->next;
        }
        reverse(x, nullptr);
        p = head;
        q = head2;
        while (p && q) {
            if (p->val != q->val) return false;
            p = p->next;
            q = q->next;
        }
        return true;
    }
};
原文地址:https://www.cnblogs.com/pk28/p/8485107.html