LeetCode-E-Palindrome Linked List

最新推荐文章于 2019-12-17 16:15:27 发布

hfsmurf

最新推荐文章于 2019-12-17 16:15:27 发布

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分类专栏：数据结构与算法文章标签： linklist palindrome

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本文介绍了一种使用双指针技巧来判断链表是否为回文的方法，并通过翻转链表的后半部分来实现O(n)的时间复杂度和O(1)的空间复杂度。

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题意

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

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解法

双指针，翻转后半段链表，和前半段做对比

实现

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