LeetCode Palindrome Linked List

题意：判断链表是否对称的,要求时间复杂度为O(n),空间复杂度为O(1)

思路：先计算链表结点数，然后将前半链表反转，然后再比较将半部和后半部是否相等

代码如下：

class ListNode
{int val;ListNode next;ListNode(int x) { val = x;}
}class Solution
{public boolean isPalindrome(ListNode head){int nodeCount = 0;ListNode p = head;while (p != null){nodeCount++;p = p.next;}ListNode current = head;ListNode pre = null;ListNode tmp = null;for (int i = 0; i < nodeCount / 2; i++){tmp = current.next;current.next = pre;pre = current;current = tmp;}if (1 == nodeCount % 2) current = current.next;for (int i = 0; i < nodeCount / 2; i++){if (pre.val != current.val) return false;else{pre = pre.next;current = current.next;}}return true;}
}

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