leetcode解题: Palindrome Linked List (234)

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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?

解法1： O（N） Time + O（1） Space

基本思路就是从中间断开，然后把其中一段reverse，再同时遍历判断两个字列是否一致。
C++

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
    
        if (!head || !head->next) {
            return true;
        }
        
        int count = 0;
        ListNode* ptr = head;
        while (ptr) {
            ++count;
            ptr = ptr->next;
        }
        
        ptr = head;
        ListNode* preMidPtr = preMid(ptr);
        ListNode* right;
        if (count % 2 == 0) {
            // even number
            right = preMidPtr->next;
            preMidPtr->next = NULL;
            
        } else {
            right = preMidPtr->next->next;
            preMidPtr->next = NULL;
        }
        
        right = reverse(right);
        while (head && right) {
            if (head->val != right->val) {
                return false;
            }
            head = head->next;
            right = right->next;
        }
        return head == NULL && right == NULL;
    }
    
    ListNode* preMid(ListNode* root) {
        ListNode* fast = root->next;
        while (fast && fast->next && fast->next->next) {
            root = root->next;
            fast = fast->next->next;
        }
        return root;
    }
    
    ListNode* reverse(ListNode* head) {
        ListNode* prev = NULL;
        while (head) {
            ListNode* temp = head->next;
            head->next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }
    
    
};

Java

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