lintcode-223-回文链表

223-回文链表

设计一种方式检查一个链表是否为回文链表。

样例

1->2->1 就是一个回文链表。

挑战

O(n)的时间和O(1)的额外空间。

标签

链表

思路

找到链表中点后，翻转链表后半部分，然后从头开始比较两个子链表。翻转链表代码源于 LintCode-35.翻转链表

code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /*
     * @param head: A ListNode.
     * @return: A boolean.
     */
    bool isPalindrome(ListNode * head) {
        // write your code here
        if (head == NULL || head->next == NULL) {
            return true;
        }
        ListNode *fast = head, *slow = head;
        while (fast->next != NULL && fast->next->next != NULL) {
            fast = fast->next->next;
            slow = slow->next;
        }

        ListNode *last = slow->next;
        slow->next = NULL;
        last = reverse(last);

        while (last != NULL) {
            if (head->val != last->val) {
                return false;
            }
            head = head->next;
            last = last->next;
        }
        if (head == NULL || head->next == NULL) {
            return true;
        }
        return false;
    }

    ListNode *reverse(ListNode *head) {
        // write your code here
        ListNode *l1 = NULL, *l2 = NULL, *l3 = NULL;

        l1 = head;
        //  链表没有节点或有一个节点 
        if (l1 == NULL || l1->next == NULL) {
            return l1;
        }
        l2 = l1->next;
        //  链表有2节点
        if (l2->next == NULL) {
            l2->next = l1;
            l1->next = NULL;
            return l2;
        }
        l3 = l2->next;
        //  链表有3个以上节点
        if (l2->next != NULL) {
            while (l2 != l3) {
                l2->next = l1;
                if (l1 == head)
                    l1->next = NULL;
                l1 = l2;
                l2 = l3;

                if (l3->next != NULL)
                    l3 = l3->next;
            }
            l2->next = l1;
            return l2;
        }
    }
};