链表的回文判断

思路:
找中间节点–>逆置->比较

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* middleNode(struct ListNode* head)
{
struct ListNode*slow=head;
struct ListNode*flast=head;
while(flast&&flast->next)
{
    slow=slow->next;
    flast=flast->next->next;
}
return slow;
}
struct ListNode* reverseList(struct ListNode* head){
    struct ListNode*newhead=NULL;
    struct ListNode*cur=head;
    while(cur)
    {
        struct ListNode*per=cur->next;
        cur->next=newhead;
        newhead=cur;
        cur=per;
    }
    return newhead;
}
bool isPalindrome(struct ListNode* head){
struct ListNode* mid=middleNode(head);//找中间节点
        struct ListNode* rmid=reverseList(mid);//逆置
        //比较
        while(head&&rmid)
        {
            if(head->val==rmid->val)
            {
                head=head->next;
                rmid=rmid->next;
            }else {
            return false;
            }
        }
        return true;
}