[LeetCode]234. Palindrome Linked List

最新推荐文章于 2025-08-10 09:12:36 发布

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最新推荐文章于 2025-08-10 09:12:36 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了解决LeetCode上234题“回文链表”的方法，通过寻找链表中点，反转前半部分，并进行逐节点比较来判断链表是否为回文结构。

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[LeetCode]234. Palindrome Linked List

题目描述

这里写图片描述

思路

首先找到中点
然后将前半段链表逆转
最后比较

代码

#include <iostream>
using namespace std;

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    bool isPalindrome(ListNode* head) {

        ListNode* slow = head;
        ListNode* fast = head;
        int size = 0;
        while (fast) {
            slow = slow->next;
            fast = fast->next;
            ++size;
            if (fast) {
                fast = fast->next;
                ++size;
            }
        }
        if (size <= 1)
            return true;
        size >>= 1;

        ListNode* pre = head;
        ListNode* cur = head->next;
        ListNode* next = head->next->next;

        pre->next = NULL;
        for (int i = 1; i < size; ++i) {
            cur->next = pre;
            pre = cur;
            cur = next;
            next = next->next;
        }
        cur = pre;
        /*
        while (cur) {
            cout << cur->val << " ";
            cur = cur->next;
        }
        cout << endl;
        */

        for (int i = 0; i < size; ++i) {
            if (cur->val != slow->val)
                return false;
            cur = cur->next;
            slow = slow->next;
        }

        return true;
    }
};

int main() {
    ListNode* l1 = new ListNode(1);
    ListNode* l2 = new ListNode(2);
    ListNode* l3 = new ListNode(1);
    ListNode* l4 = new ListNode(2);
    ListNode* l5 = new ListNode(1);

    l1->next = l2;
    l2->next = l3; 
    l3->next = l4;
    l4->next = l5;
    Solution s;
    cout << s.isPalindrome(l1) << endl;

    system("pause");
    return 0;
}