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

Example 1:

Input: 1->2Output: false

Example 2:

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

代码：

/** * 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) return true;        string s = "";        ListNode *cur = head;        while(cur) {            s += (cur -> val) + '0';            cur = cur -> next;        }        string ss = s;        reverse(s.begin(), s.end());        if(ss == s) return true;        return false;    }};

　　妈耶 自己写对的第一个链表 也太开心了 8！！！！！！！ 今日份的七彩开心