LeetCode | Palindrome Linked List

https://leetcode.com/problems/palindrome-linked-list/

思路1：遍历一次链表，用一个数组存储链表节点值，然后用双指针法判断数组是否是回文的。需要额外O(n)的空间。

C++

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

思路2：如果不使用额外空间的话，就必须修改输入链表了。先找到链表中点，然后将右半部分链表反转，再判断左右两个链表节点值是否相等。

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if (!head || !head->next) return true;
        
        // find middle node
        ListNode *slow = head, *fast = head->next;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode *l = head, *r = slow->next;
        slow->next = NULL;  // cut
        
        // reverse r
        r = reverse_list(r);
        
        // is equal
        while (l && r) {
            if (l->val != r->val) return false;
            l = l->next;
            r = r->next;
        }
        return true;
    }
    
    ListNode* reverse_list(ListNode* head) {
        ListNode *prev = NULL;
        while (head) {
            ListNode *next = head->next;
            head->next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
};