[Leetcode]234. Palindrome Linked List(C++)

题目描述

题目链接：234. Palindrome Linked List

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

例子

例子 1

Input: 1->2 Output: false

例子 2

Input: 1->2->2->1 Output: true

Follow Up

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

解题思路

题目要求判断链表是否回文（顺着读和反着读顺序数字一样），并且要求线性时间和常数空间内完成。我们可以首先计算出链表的长度，然后从中间断开链表，（如果是奇数个节点不考虑中间那个节点），再将后半段链表反转一下，最后依次比较两半链表即可。代码如下：

 * 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) {

        // compute the length of linked list
        int n = 0;
        ListNode *curr = head;
        while(curr) {
            curr = curr->next;
            n++;
        }
        if (n <= 1) return true;

        // move curr to the second half of linked list
        curr = head;
        for (int i = 0; i < n / 2; i++) {
            curr = curr->next;
        }
        if (n % 2) {
            curr = curr->next;
        }

        // reverse the second half of linked list
        ListNode *prev = nullptr;
        ListNode *next = curr->next;
        curr->next = prev;
        while(next) {
            prev = curr;
            curr = next;
            next = curr->next;
            curr->next = prev;
        }

        // compare the first half & reversed second half
        while(curr && head) {
            if (curr->val != head->val) return false;
            curr = curr->next;
            head = head->next;
        }

        return true;
    }
};

时间复杂度: O(n)
空间复杂度: O(1)

GitHub 代码同步地址： 234.PalindromeLinkedList.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions