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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?
思路1：创建一个ArrayList用于保存每个节点的value，遍历LinkedList将每个节点的value依次add到ArrayList中。
利用双指针，一个在ArrayList的首部，一个在ArrayList的尾部，相向遍历看其value是否相等，如果不相等返回false，如果相遇则返回true。
其空间复杂度为O(n),时间复杂度为O(1).
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public boolean isPalindrome(ListNode head) { if (head == null || head.next == null) { return true; } ArrayList<Integer> list = new ArrayList<>(); while (head != null) { list.add(head.val); head = head.next; } for (int i = 0, j = list.size() - 1; i < j; i++, j--) { if ((int)list.get(i) != (int)list.get(j)) { return false; } } return true; } }
思路2：将链表从中间节点之后的节点都翻转记为p2, 原链表到中间节点的部分记为p1.
依次比较p1 与 p2 节点值是否相等。
1 // This code would destroy the original structure of the linked list. 2 // If you do not want to destroy the structure, you can reserve the second part back. 3 public class Solution { 4 /** 5 * @param head a ListNode 6 * @return a boolean 7 */ 8 public boolean isPalindrome(ListNode head) { 9 if (head == null) { 10 return true; 11 } 12 13 ListNode middle = findMiddle(head); 14 middle.next = reverse(middle.next); 15 16 ListNode p1 = head, p2 = middle.next; 17 while (p1 != null && p2 != null && p1.val == p2.val) { 18 p1 = p1.next; 19 p2 = p2.next; 20 } 21 22 return p2 == null; 23 } 24 25 private ListNode findMiddle(ListNode head) { 26 if (head == null) { 27 return null; 28 } 29 ListNode slow = head, fast = head.next; 30 while (fast != null && fast.next != null) { 31 slow = slow.next; 32 fast = fast.next.next; 33 } 34 35 return slow; 36 } 37 38 private ListNode reverse(ListNode head) { 39 ListNode prev = null; 40 41 while (head != null) { 42 ListNode temp = head.next; 43 head.next = prev; 44 prev = head; 45 head = temp; 46 } 47 48 return prev; 49 } 50 }
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原文地址：http://www.cnblogs.com/FLAGyuri/p/5314003.html
