回文链表

请判断一个链表是否为回文链表。

你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？

 

（1）用数组存储，然后用数组比较。

/**
 * 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;
        }
        vector<int> vec;
        ListNode *r = head;
        int len = 0;
        while(r != NULL){
            vec.push_back(r->val);
            r = r->next;
            len++;
        }
        for(int i = 0; i < len/2; i++){
            if(vec[i] != vec[len-1-i]){
                return false;
            }
        }
        return true;
    }
};

 

 

（2）递归

class Solution {
public:
    ListNode *f;
    bool isPalindrome(ListNode* head) {
        if(head == NULL || head->next == NULL){
            return true;
        }
        f = head;
        return check(head);
    }
    
    bool check(ListNode *p){
        if(p->next != NULL){
            if(!check(p->next)){
                return false;
            }
            if(f->val != p->val){
                return false;
            }
        }
        f = f->next;
        return true;
    }
};

 

 

（3）双指针找中间+反转后半部分

class Solution {
public:
    ListNode *f;
    bool isPalindrome(ListNode* head) {
        if(head == NULL || head->next == NULL){
            return true;
        }
        
        ListNode *s = head;
        ListNode *f = head->next;
        while(f != NULL && f->next != NULL){
            s = s->next;
            f = f->next->next;
        }
        
        ListNode *reverseHead = reverseList(s->next);
        
        s = head;
        f = reverseHead;
        while(f != NULL){
            if(s->val != f->val){
                return false;
            }
            s = s->next;
            f = f->next;
        }
        return true;
    }
    
    ListNode* reverseList(ListNode* head) {
        if(head != NULL){
            ListNode *p = head->next;
            ListNode *q = head;
            while(p != NULL){
                q->next = p->next;
                p->next = head;
                head = p;
                p = q->next;
            }
        }
        return head;
    }
};