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程式語言 - LeetCode - C++ - 234. Palindrome Linked List

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題目：


方法：

1. 找中點(快慢指標)
2. 反轉後半段linked list
3. 比較前半 vs 後半

解答：

/**
 * 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 {
public:
    bool isPalindrome(ListNode* head) {
        ListNode *fast = head;
        ListNode *slow = head;

        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }

        ListNode* pre = nullptr;
        while (slow) {
            ListNode *next = slow->next;
            slow->next = pre;
            pre = slow;
            slow = next;
        }

        while (pre) {
            if (pre->val != head->val) {
                return false;
            }

            head = head->next;
            pre = pre->next;
        }

        return true;
    }
};