检测链表是否是palindrome.
思路1:翻转并比较。
思路2:迭代。
思路3:递归。
public static boolean isPalindrome(LinkedListNode head) { LinkedListNode fast = head; LinkedListNode slow = head; Stack<Integer> stack = new Stack<Integer>(); while (fast != null && fast.next != null) { stack.push(slow.data); slow = slow.next; fast = fast.next.next; } /* Has odd number of elements, so skip the middle */ if (fast != null) { slow = slow.next; } while (slow != null) { int top = stack.pop().intValue(); System.out.println(slow.data + " " + top); if (top != slow.data) { return false; } slow = slow.next; } return true; } public Question.Result isPalindromeRecurse(LinkedListNode head, int length) { if (head == null || length == 0) { return new Question.Result(null, true); } else if (length == 1) { return new Question.Result(head.next, true); } else if (length == 2) { return new Question.Result(head.next.next, head.data == head.next.data); } Question.Result res = isPalindromeRecurse(head.next, length - 2); if (!res.result || res.node == null) { return res; // Only "result" member is actually used in the call stack. } else { res.result = head.data == res.node.data; res.node = res.node.next; return res; } } public boolean isPalindrome(LinkedListNode head) { int size = 0; LinkedListNode n = head; while (n != null) { size++; n = n.next; } Result p = isPalindromeRecurse(head, size); return p.result; }
Cracking the Coding Interview Q2.7,布布扣,bubuko.com
Cracking the Coding Interview Q2.7
原文:http://www.cnblogs.com/jdflyfly/p/3830700.html