Palindrome Linked List

最新推荐文章于 2021-03-02 15:23:07 发布

原创最新推荐文章于 2021-03-02 15:23:07 发布 · 268 阅读

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本文介绍了一种高效判断单链表是否为回文的方法。通过使用快慢指针找到链表的中点，并反转后半部分链表，进而逐个比较节点值来判断链表是否为回文结构。

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

注解：用快慢指针找出中间节点，将后半链表倒序，对比每个节点

优化：找中间节点时，翻转前半段链表

bool isPalindrome(ListNode* head) {
    if( !head || !head->next )
        return true;
    ListNode *fast = head, *slow = head;
    while( fast->next && fast->next->next )
    {
        slow = slow->next;
        fast = fast->next->next;
    }
    
    if( fast->next != NULL )
    {
        fast = slow->next;
        slow = reverse( head, fast );
    }
    else
    {
        fast = slow->next;
        slow = reverse( head, slow );
    }
    while( slow && fast )
    {
        if( slow->val != fast->val )
            return false;
        slow = slow->next;
        fast = fast->next;
    }
    
    return true;
}
    
ListNode* reverse( ListNode* head, ListNode* tail )
{
    if( head == tail || head->next == tail )
        return head;
    
    ListNode *p = head->next, *q;
    head->next = NULL;
    while( p != tail )
    {
        q = p->next;
        p->next = head;
        head = p;
        p = q;
    }
    return head;
}