LeetCode Algorithm 剑指 Offer II 027. 回文链表

题目链接：剑指 Offer II 027. 回文链表

Ideas

算法：双指针
数据结构：链表
思路：首先把链表的后半段翻转过来，然后用两个指针分别从头和从中间开始遍历判断是否相同，如果相同则是回文链表，否则不是回文链表，但不管是不是回文链表，翻转的链表还是要翻转回去。
1.用快慢指针，快指针quick一次走两步，慢指针slow一次走一步，当快指针到头时，慢指针正好到链表中间；（如果链表长度为奇数，快指针最终走到最后一个元素，如果链表长度为偶数，快指针最终走到倒数第二个元素）
2.从链表中间开始，将后面的节点逐个翻转，即当前节点原本指向下一个节点，改为指向前一个节点；
3.双指针分别从链表的头尾出发，遍历判断当前节点值是否相等，最终双指针相遇时截止；
4.从链表中间开始，再将后面的节点翻转回去。

Code

C++

class Solution {
    
public:
    bool isPalindrome(ListNode* head) {
    
        if (head == nullptr || head->next == nullptr) {
    
            return true;
        }
        // 1.快慢指针找到链表中间位置
        ListNode *quick = head, *slow = head;
        while (quick->next != nullptr && quick->next->next != nullptr) {
    
            slow = slow->next;
            quick = quick->next->next;
        }
        // 2.从链表中间位置开始将后续节点翻转
        ListNode *cur = slow->next;
        slow->next = nullptr;
        while (cur != nullptr) {
    
            ListNode *nxt = cur->next;
            cur->next = slow;
            slow = cur;
            cur = nxt;
        }
        // 3.双指针分别从头尾开始遍历
        bool res = true;
        quick = head;
        ListNode *tail = slow;
        while (quick != nullptr && slow != nullptr) {
    
            if (quick->val != slow->val) {
    
                res = false;
                break;
            }
            quick = quick->next;
            slow = slow->next;
        }
        // 4.将翻转的节点再翻转回去
        cur = tail->next;
        tail->next = nullptr;
        while (cur != nullptr) {
    
            ListNode *nxt = cur->next;
            cur->next = tail;
            tail = cur;
            cur = nxt;
        }
        return res;
    }
};

当然，如果你觉得双指针比较麻烦的话，也可以遍历链表保存到数组中再判断是否为回文。

class Solution {
    
public:
    bool isPalindrome(ListNode* head) {
    
		vector<int> values;
		while (head) {
    
			values.push_back(head->val);
			head = head->next;
		}
		for (int i = 0, j = values.size() - 1; i < j; i++, j--) {
    
			if (values[i] != values[j]) {
    
				return false;
			}
		}
		return true;
    }
};

Python

class Solution:
    def findFirstHalfEnd(self, node):
        fast = slow = node
        while fast.next is not None and fast.next.next is not None:
            fast = fast.next.next
            slow = slow.next
        return slow

    def reverseList(self, node):
        previous, current = None, node
        while current is not None:
            nextNode = current.next
            current.next = previous
            previous = current
            current = nextNode
        return previous

    def isPalindrome(self, head: ListNode) -> bool:
        # 判断边界情况
        if head is None:
            return True

        # 找到前半部分链表的为节点并反转后半部分链表
        firstHalfEnd = self.findFirstHalfEnd(head)
        secondHalfStart = self.reverseList(firstHalfEnd.next)

        # 判断是否为回文
        ans, firstPosition, secondPosition = True, head, secondHalfStart
        while ans and secondPosition is not None:
            if firstPosition.val != secondPosition.val:
                return False
            firstPosition = firstPosition.next
            secondPosition = secondPosition.next

        # 还原反转的链表
        firstHalfEnd.next = self.reverseList(secondHalfStart)

        return ans