LeetCode | Linked List Cycle II

题目

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Follow up:
Can you solve it without using extra space?

回环检测问题，参考维基百科http://en.wikipedia.org/wiki/Cycle_detection，发现有两种解法。

一种的Floyd提出的经典的龟兔（Tortoise and Hare）算法，另一种是Brent提出的改进算法，性能上有明显的提升。

通过LeetCode测试结果来看，Brent提出的算法性能确实更好。

代码

public class LinkedListCycleII {
	class ListNode {
		int val;
		ListNode next;

		ListNode(int x) {
			val = x;
			next = null;
		}
	}

	public ListNode detectCycle(ListNode head) {
		// return floydAlgorithm(head);
		return brentAlgorithm(head);
	}

	private ListNode brentAlgorithm(ListNode head) {
		if (head == null || head.next == null) {
			return null;
		}

		int power = 1;
		int lam = 1;

		ListNode tortoise = null;
		ListNode hare = head;
		while (hare != tortoise) {
			if (hare == null) {
				return null;
			}
			if (power == lam) {
				tortoise = hare;
				power *= 2;
				lam = 0;
			}
			hare = hare.next;
			++lam;
		}

		hare = head;
		tortoise = head;
		for (int i = 0; i < lam; ++i) {
			hare = hare.next;
		}
		while (hare != tortoise) {
			tortoise = tortoise.next;
			hare = hare.next;
		}

		return tortoise;
	}

	private ListNode floydAlgorithm(ListNode head) {
		if (head == null || head.next == null) {
			return null;
		}

		ListNode hare = head.next.next;
		ListNode tortoise = head.next;
		while (hare != tortoise) {
			if (hare == null || hare.next == null) {
				return null;
			}
			hare = hare.next.next;
			tortoise = tortoise.next;
		}

		tortoise = head;
		while (tortoise != hare) {
			hare = hare.next;
			tortoise = tortoise.next;
		}

		return tortoise;
	}
}

LeetCode | Linked List Cycle II

原文：http://blog.csdn.net/perfect8886/article/details/19092179