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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?
Analyse: Obtain the elements first, then judge.
Runtime: 28ms.
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 bool isPalindrome(ListNode* head) { 12 if(!head || !head->next) return true; 13 14 vector<int> v; 15 while(head){ 16 v.push_back(head->val); 17 head = head->next; 18 } 19 for(int i = 0, j = v.size() - 1; i < j; i++, j--){ 20 if(v[i] != v[j]) return false; 21 } 22 return true; 23 } 24 };
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原文地址:http://www.cnblogs.com/amazingzoe/p/4777521.html