LeetCode 234. Palindrome Linked List (回文链表)
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?
题目标签:Linked List
题目给了我们一个 linked list,让我们判断它是不是回文。
这里可以利用 #206 Reverse Linked List 把右边一半的链表 倒转,然后从左右两头开始比较链表是否是回文。
这样的话,首先要找到链表的中间点,然后开始倒转右半边链表。
可以利用slow fast pointers 来找到中间点,当fast 走完的时候,slow 正好在中间。
Java Solution:
Runtime beats 39.01%
完成日期:06/11/2017
关键词:singly-linked list
关键点:利用slow fast 指针来找到中间点
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 class Solution 10 { 11 public boolean isPalindrome(ListNode head) 12 { 13 ListNode slow = head; 14 ListNode fast = head; 15 ListNode tail; 16 17 // find the middle 18 while(fast != null && fast.next != null) 19 { 20 slow = slow.next; 21 fast = fast.next.next; 22 } 23 24 if(fast != null) // odd length 25 slow = slow.next; // move middle to right half 26 27 // reverse right half 28 tail = reverse(slow); 29 30 // compare head and tail 31 while(tail != null) 32 { 33 if(head.val != tail.val) 34 return false; 35 36 head = head.next; 37 tail = tail.next; 38 } 39 40 return true; 41 } 42 43 private ListNode reverse(ListNode cursor) 44 { 45 ListNode pre = null; 46 ListNode next; 47 48 while(cursor != null) 49 { 50 next = cursor.next; 51 cursor.next = pre; 52 pre = cursor; 53 cursor = next; 54 } 55 56 return pre; // tail node 57 } 58 }
参考资料:https://discuss.leetcode.com/topic/33376/java-easy-to-understand
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