234. Palindrome Linked List (Easy)

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

 

Follow up: Could you do it in O(n) time and O(1) space?

A:

就是简单的读2遍，然后一个从后往前，一个从前往后。

/**
 * 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) {
        // iterate twice
        vector<int> V;
        auto runner = head;
        while (runner) {
            V.push_back(runner->val);
            runner = runner->next;
        }
        runner = head;
        while (runner) {
            auto last = V.back();
            V.pop_back();
            if (last != runner->val) {
                return false;
            }
            runner = runner->next;
        }
        return true;
    }
};

Follow up:

Could you do it in O(n) time and O(1) space?

1：  reverse后半段。

2： 利用 recursive  的call stack,来对比