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

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

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刚开始没住时间复杂度和空间复杂度：

/** * 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) {        ListNode* tempHead = head;        int count = 0;        while(tempHead)        {         ++count;         tempHead = tempHead->next;        }        stack<ListNode*> tempStack;        tempHead = head;        for(int i = 1; i <= count/2; ++i)        {            tempStack.push(tempHead);            tempHead = tempHead->next;        }        if(count % 2 )            tempHead = tempHead->next;        while(tempHead)        {            if(tempStack.top()->val != tempHead->val)                return false;            else                {                    tempStack.pop();                    tempHead = tempHead->next;                }        }        return true;    }};

如果要求时间和空间复杂度：

将前面一般逆序，再比。空间复杂度为0（1）

/** * 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( NULL == head || NULL == head->next )            return true;        ListNode* tempHead = head;        int count = 0;        while(tempHead)        {         ++count;         tempHead = tempHead->next;        }        ListNode* pre = NULL;        ListNode* nxt = head;        for(int i = 1; i <= count/2; ++i)        {           tempHead = nxt->next;           nxt->next = pre;           pre = nxt;           nxt = tempHead;        }        if(count % 2 )            nxt = nxt->next;        while(nxt)        {            if(nxt->val != pre->val)                return false;            else            {                pre = pre->next;                nxt = nxt->next;            }        }                return true;                        /*        stack<ListNode*> tempStack;        tempHead = head;        for(int i = 1; i <= count/2; ++i)        {            tempStack.push(tempHead);            tempHead = tempHead->next;        }        if(count % 2 )            tempHead = tempHead->next;        while(tempHead)        {            if(tempStack.top()->val != tempHead->val)                return false;            else                {                    tempStack.pop();                    tempHead = tempHead->next;                }        }        return true;*/    }};

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