palindrome-linked-list

234. Palindrome Linked List

Given the head of a singly linked list, return true_ if it is a _

palindrome

_ or false otherwise_.

Example 1:

Input: head = [1,2,2,1] Output: true

Example 2:

Input: head = [1,2] Output: false

Constraints:

The number of nodes in the list is in the range [1, 105].
0 <= Node.val <= 9

判斷 linked list 是否有迴文

解法一：遞迴

/**
 * 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* cur = head;
        return helper(head, cur);
    }
    bool helper(ListNode* head, ListNode*& cur) {
        if (!head) return true;
        bool res = helper(head->next, cur) && (cur->val == head->val);
        cur = cur->next;
        return res;
    }
};

T: $O(N)$
S: $O(N)$

/**
 * 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) {
        if (!head || !head->next) return true;
        ListNode* slow = head;
        ListNode* fast = head;
        stack<int> stk;
        stk.push(head->val);
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
            stk.push(slow->val);
        }
        if (!fast->next) stk.pop();

        while (slow && slow->next) {
            slow = slow->next;
            int tmp = stk.top(); stk.pop();
            if (tmp != slow->val) return false;
        }
        return true;
    }
};

T: $O(N)$
S: $O(N)$

不用 stack，空間複雜度 $O(1)$ 的解法

T: $O(N)$
S: $O(1)$

234. Palindrome Linked List​

234. Palindrome Linked List