回文链表

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快慢指针找中点，反转后半段，与前半段比较，O(n) O(1)。

OJ: leetcodecn

题目 ID: palindrome-linked-list

难度:入门

标签:链表双指针栈cpppython

日期: 2026-07-28 22:05

目录

  1. 题意
  2. 思路
  3. 代码
  4. 复杂度
  5. 总结

题意

判断单链表是否为回文。

思路

数组存值再双指针 O(n) 空间。优化：快慢指针找到中间节点，反转后半段，逐节点比较。

代码

cpp
/**
 * Author by Rainboy
 */
// main.cpp：快慢指针找中点，反转后半段，比较。
#include <bits/stdc++.h>
using namespace std;

struct ListNode {
    int val;
    ListNode *next;

    ListNode(int x) : val(x), next(nullptr) {}
};

class Solution {
public:
    bool isPalindrome(ListNode *head) {
        if (!head || !head->next)
            return true;
        auto slow = head, fast = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        auto mid = slow->next;
        slow->next = nullptr;
        ListNode *prev = nullptr;
        while (mid) {
            auto nxt = mid->next;
            mid->next = prev;
            prev = mid;
            mid = nxt;
        }
        auto a = head, b = prev;
        while (b) {
            if (a->val != b->val)
                return false;
            a = a->next;
            b = b->next;
        }
        return true;
    }
};

ListNode *build(istream &in, int n) {
    if (!n)
        return nullptr;
    auto head = new ListNode(0), cur = head;
    for (int i = 0, v; i < n; i++) {
        in >> v;
        cur->next = new ListNode(v);
        cur = cur->next;
    }
    return head->next;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    auto head = build(cin, n);
    cout << Solution().isPalindrome(head) << '\n';
    return 0;
}
python
#!/usr/bin/env python3
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def isPalindrome(self, head: ListNode) -> bool:
        if not head or not head.next:
            return True
        slow = fast = head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        mid = slow.next
        slow.next = None
        prev = None
        while mid:
            nxt = mid.next
            mid.next = prev
            prev = mid
            mid = nxt
        a, b = head, prev
        while b:
            if a.val != b.val:
                return False
            a, b = a.next, b.next
        return True


def build(arr):
    dummy = cur = ListNode(0)
    for v in arr:
        cur.next = ListNode(v)
        cur = cur.next
    return dummy.next


def main() -> None:
    n = int(input())
    a = list(map(int, input().split()))
    head = build(a)
    print(Solution().isPalindrome(head))


if __name__ == "__main__":
    main()

复杂度

  • 时间复杂度：O(n)。
  • 空间复杂度：O(1)。

总结

链表中点 + 反转是回文判断的标准做法。注意奇偶长度下中点的定位。