[C语言][LeetCode][234]Palindrome Linked List

题目

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?

标签

Linked List、Two Pointers

难度

简单

分析

题目意思是判断一个链表是否是回文结构。做法是先拿到链表的中间元素，然后逆序后面一段的链表，最后比较前段和后段的链表。

C代码实现

struct ListNode* linked_list_find_mid_ele(struct ListNode* head)
{
    struct ListNode* fast, *slow;

    fast = head;
    slow = head;

    if(!head)
        return NULL;

    while(fast && fast->next)
    {
        fast = fast->next->next;
        slow = slow->next;
    }

    return slow;
}

struct ListNode* reverseList(struct ListNode* head) 
{
    struct ListNode* t;
    struct ListNode* p, * q;

    if(!head || !head->next)
        return head;

    p = head; 
    q = head->next;

    while(q)
    {
        t = q->next;
        q->next = p;
        p = q;
        q = t;
    }

    head->next = NULL;
    head = p;

    return head;
}

bool isPalindrome(struct ListNode* head) 
{
    struct ListNode * midNode;
    struct ListNode *leftList, *rightList;

    if(!head || !head->next)
        return true;

    leftList = head;    
    midNode = linked_list_find_mid_ele(head);

    rightList = reverseList(midNode);

    while(leftList && rightList)
    {
        if(leftList->val != rightList->val)
            return false;
        leftList = leftList->next;
        rightList = rightList->next;
    }

    return true;
}