leetcode 234: Palindrome Linked List

最新推荐文章于 2024-11-10 17:00:32 发布

mikefan1991

最新推荐文章于 2024-11-10 17:00:32 发布

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本文介绍了一种使用O(1)空间复杂度判断单链表是否为回文的方法。通过快慢指针找到中间节点并反转后半部分，再进行比较实现。此算法避免了额外空间的使用。

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To solve it with only O(1) space, you can only find the mid point of the linked list and do some reverse operation, so that you are able to traverse back. To find the mid point, use the fast and slow method, and once the mid point is found, reverse the latter half and compare.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode *slow,*fast;
        slow=fast=head;
        while(fast&&fast->next)
        {
            slow=slow->next;
            fast=fast->next->next;
        }
        reverse(slow,fast);
        slow=head;
        while(fast)
        {
            if(slow->val!=fast->val)
                return false;
            slow=slow->next;
            fast=fast->next;
        }
        return true;
    }
    ListNode* reverse(ListNode* root,ListNode*& end)
    {
        if(!root||!root->next)
        {
            end=root;
            return root;
        }
        ListNode* temp=reverse(root->next,end);
        temp->next=root;
        root->next=NULL;
        return root;
    }
};