【leetcode】【easy】234. Palindrome Linked List

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

Example 1:

Input: 1->2
Output: false
Example 2:

Input: 1->2->2->1
Output: true
Follow up:
Could you do it in O(n) time and O(1) space?

题目链接：https://leetcode-cn.com/problems/palindrome-linked-list/

 

思路

题目要求时间复杂度O(n)，说明不能用暴力法；

要求空间复杂度O(1)，说明不能额外开空间储存，比如不能用栈。

因此确定思路为：先用快慢指针找中点，将后半子链表逆序，最后依次比较两个子链的元素。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if(head==NULL || head->next==NULL) return true;
        auto l = head, r = head;
        while(r->next){
            r = r->next;
            if(r->next){
                r = r->next;
                l = l->next;
            }
        }
        r = l->next;
        auto nxt = r->next;
        while(nxt){
            auto last = r;
            r = nxt;
            nxt = r->next;
            r->next = last;
        }
        l->next->next = NULL;
        l->next = NULL;
        l = head;
        while(r){
            if(l->val!=r->val) return false;
            else{
                l = l->next;
                r = r->next;
            }
        }
        return true;
    }
};