234. Palindrome Linked List

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  • Difficulty: Easy
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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?

分析

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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 == NULL || head->next == NULL) return true;
         
        //use slow and fast pointer to find the mid node
        ListNode* slow = head, * fast = head;
        while(fast && fast->next && fast->next->next){
            slow = slow->next;
            fast = fast->next->next;
        }
         
        //reverse the right half list
        ListNode* l1 = head;
        ListNode* l2 = slow->next;
        slow->next = NULL;
        l2 = reverse(l2);
         
        //compare the two lists's node vale
        while(l2){
            if(l1->val != l2->val) return false;
            l1 = l1->next;
            l2 = l2->next;
        }
         
        return true;
    }
     
    ListNode* reverse(ListNode* head){
        ListNode* pre = NULL;
        ListNode* cur = head;
        while(cur){
            ListNode* tmp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = tmp;
        }
        return pre;
    }
};




posted @ 2017-02-20 15:42  copperface  阅读(169)  评论(0)  编辑  收藏  举报