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简单
栈
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链表
双指针

234. 回文链表

English Version

题目描述

给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。

 

示例 1：

输入：head = [1,2,2,1]
输出：true

示例 2：

输入：head = [1,2]
输出：false

 

提示：

  • 链表中节点数目在范围[1, 105] 内
  • 0 <= Node.val <= 9

 

进阶：你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？

解法

方法一：快慢指针

我们可以先用快慢指针找到链表的中点，接着反转右半部分的链表。然后同时遍历前后两段链表，若前后两段链表节点对应的值不等，说明不是回文链表，否则说明是回文链表。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为链表的长度。

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        slow, fast = head, head.next
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
        pre, cur = None, slow.next
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        while pre:
            if pre.val != head.val:
                return False
            pre, head = pre.next, head.next
        return True

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        while (pre != null) {
            if (pre.val != head.val) {
                return false;
            }
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head->next;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode* pre = nullptr;
        ListNode* cur = slow->next;
        while (cur) {
            ListNode* t = cur->next;
            cur->next = pre;
            pre = cur;
            cur = t;
        }
        while (pre) {
            if (pre->val != head->val) return false;
            pre = pre->next;
            head = head->next;
        }
        return true;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func isPalindrome(head *ListNode) bool {
	slow, fast := head, head.Next
	for fast != nil && fast.Next != nil {
		slow, fast = slow.Next, fast.Next.Next
	}
	var pre *ListNode
	cur := slow.Next
	for cur != nil {
		t := cur.Next
		cur.Next = pre
		pre = cur
		cur = t
	}
	for pre != nil {
		if pre.Val != head.Val {
			return false
		}
		pre, head = pre.Next, head.Next
	}
	return true
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function isPalindrome(head: ListNode | null): boolean {
    let slow: ListNode = head,
        fast: ListNode = head.next;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
    }
    let cur: ListNode = slow.next;
    slow.next = null;
    let prev: ListNode = null;
    while (cur != null) {
        let t: ListNode = cur.next;
        cur.next = prev;
        prev = cur;
        cur = t;
    }
    while (prev != null) {
        if (prev.val != head.val) return false;
        prev = prev.next;
        head = head.next;
    }
    return true;
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {boolean}
 */
var isPalindrome = function (head) {
    let slow = head;
    let fast = head.next;
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    let cur = slow.next;
    slow.next = null;
    let pre = null;
    while (cur) {
        let t = cur.next;
        cur.next = pre;
        pre = cur;
        cur = t;
    }
    while (pre) {
        if (pre.val !== head.val) {
            return false;
        }
        pre = pre.next;
        head = head.next;
    }
    return true;
};

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public bool IsPalindrome(ListNode head) {
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        while (pre != null) {
            if (pre.val != head.val) {
                return false;
            }
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}