234. Palindrome Linked List

/*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?判断一个单链表是否为回文 思路1：     把链表后半部分反序排列，然后从链表头部与后半部分开始位置比较即可。思路2：利用递归函数的栈处理方式，递归到链表的最后一个节点与第一个节点比较，然后向中间节点移动。*//** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */bool isPalindrome(struct ListNode* head) {    if(!head || !head->next) return 1;struct ListNode *fast=head,*slow=head;while(fast->next && fast->next->next)fast=fast->next->next,slow=slow->next;struct ListNode *secondHead=slow->next;reverseList(&secondHead);while(secondHead){if(head->val != secondHead->val)return 0;head=head->next;secondHead=secondHead->next;}return 1;}void reverseList(struct ListNode **head){if(!(*head) || !(*head)->next) return ;struct ListNode **h=head;struct ListNode *cur=*h,*pre=NULL;pre=cur->next;while(pre){cur->next=pre->next;pre->next=*h;*h=pre;pre=cur->next;}}