【easy】234. Palindrome Linked List

ques: 判断一个链表是否回文 Could you do it in O(n) time and O(1) space?

method：先将链表分为两部分，将后半部分反转，最后从前往后判断是否相等。

topic: 链表，链表反转

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */  

class Solution {
public:
     bool isPalindrome(ListNode *head) {
        //input check  abcba abccba
        if(head == NULL || (head&&head->next==NULL)) return true;  

        ListNode *middle = partition(head);
        middle = reverse(middle);  

        while(head!=NULL && middle!=NULL) {
            if(head->val != middle->val) return false;
            head = head->next;
            middle = middle->next;
        }
        return true;
    }
private:
    ListNode* partition(ListNode *head) {
        ListNode* p = head;
        while(p&&p->next!=NULL && p->next->next!=NULL) {
            p = p->next->next;
            head = head->next;
        }  

        p = head->next;
        head->next = NULL;
        return p;
    }
private:
    ListNode* reverse(ListNode *head) {
        if(head==NULL || head->next==NULL) return head;
        ListNode* pre = head;
        ListNode* cur = head->next;
        pre->next = NULL;
        ListNode* nxt = NULL;  

        while(cur!=NULL) {
            nxt = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nxt;
        }
        return pre;
    }
};