题解 | #判断一个链表是否为回文结构#

双链表法解决，将原链表的后半段翻转。

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 the head
 * @return bool布尔型
 */
#include <stdlib.h>
//将后半段数据翻转
struct ListNode* reverse_halfln(struct ListNode* head){
    struct ListNode* resl = head;
    struct ListNode* p = head;
    struct ListNode* q = head->next;
    while(q->next!=NULL && q!=NULL){
        p = p->next;
        q = q->next->next;
    }
    //start为后半段 
    struct ListNode* start = p->next;
    struct ListNode* start_n = start->next;
    struct ListNode* temp_ln = (struct ListNode*)malloc(sizeof(struct ListNode));
    temp_ln->next = NULL; 
    while(start!=NULL){
        start->next = temp_ln->next;
        temp_ln->next = start;
        start = start_n;
        if(start != NULL){
            start_n = start ->next;
        }
    }
    //防止断链
    p->next = temp_ln->next;
    return resl;
}
bool isPail(struct ListNode* head ) {
    // write code here
    if(head->next == NULL) return true;
    //翻转后半段得到新的链表
    struct ListNode* new_ln = reverse_halfln(head);
    struct ListNode* p = new_ln;
    struct ListNode* q = new_ln->next;
    while(q->next!=NULL && q!=NULL){
        p = p->next;
        q = q->next->next;
    }
    //p为后半段开头，q为前半段开头
    p = p->next;
    q = new_ln;
    while(p!=NULL){
        if(p->val!=q->val)
            return false;
        p = p->next;
        q = q->next;
    }
    return true;
    
    //return true;
}