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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?
public bool IsPalindrome(ListNode head) { if(head == null) return true; // Reverse the head; ListNode standForward = new ListNode(0); ListNode stand =standForward; ListNode newHead = null; while(head != null) { standForward.next = new ListNode(head.val); ListNode nextNode = head.next; head.next = newHead; newHead = head; head = nextNode; standForward = standForward.next; } while(newHead != null) { if(stand.next.val != newHead.val) return false; stand=stand.next ; newHead = newHead.next; } return true; }
if use space O(1), we need to use 2 pointer to find the middle pointer of linked list and then reverse the second part of the linked list.
public bool IsPalindrome(ListNode head) { if(head == null) return true; ListNode slow = head; ListNode fast = head; ListNode nextNode = null; ListNode newHead = null; //get middle pointer of the linked list while(fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } while(slow != null) { nextNode = slow.next; slow.next = newHead; newHead = slow; slow = nextNode; } while(head != null && newHead != null) { if(head.val != newHead.val) return false; head = head.next; newHead = newHead.next; } return true; }
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原文地址：http://www.cnblogs.com/renyualbert/p/5854771.html
