题目浅析

  • 想查看原题可以点击题目链接。

  • 简单地说，就是规定回文链表就是向前和向后读都相同序列的链表，求判断当前链表是否是回文链表。

思路分享

代码解答（强烈建议自行解答后再看）

  • 参考题解
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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def get_mid(self, head):
        left, right = head, head
        while right and right.next:
            left, right = left.next, right.next.next
        return left
    
    def reserve_list(self, head):
        pre = None
        cur = head
        while cur:
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        return pre
    
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        mid = self.get_mid(head)
        mid = self.reserve_list(mid)
        while mid:
            if head.val == mid.val:
                # print(head, mid) # 虽然前一段的最后一个指针没有编辑，但不妨碍答案的检测
                head, mid = head.next, mid.next
            else:
                return False
        return True