LeetCode Linked List Cycle II 和I 通用算法和优化算法

Linked List Cycle II

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?

和问题一Linked List Cycle几乎一样。如果用我的之前的解法的话，可以很小修改就可以实现这道算法了。但是如果问题一用优化了的解法的话，那么就不适用于这里了。下面是我给出的解法，可以看得出，这里需要修改很小地方就可以了。 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:

	bool find(ListNode *head, ListNode *testpNode)
	{
		ListNode *p = head;
		while (p != testpNode->next)
		{
			if(p == testpNode)
				return false;
			p = p->next;
		}
		return true;
	}


	ListNode *detectCycle(ListNode *head) {
		// IMPORTANT: Please reset any member data you declared, as
		// the same Solution instance will be reused for each test case.
		if(head == NULL)
			return false;

		ListNode *cur = head;
		while(cur != NULL)
		{
			if(find(head, cur))
				return cur->next;
			cur = cur->next;
		}
		return NULL;
	}
};

然后转一下下面那位朋友的博客，他的解法很优化，不过只适合第一个LeetCode Linked List Cycle问题，而不适合这里。值得学习学习，一起贴在这里了。

http://blog.csdn.net/doc_sgl/article/details/13614853

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ListNode* pfast = head;
		ListNode* pslow = head;
		do{
			if(pfast!=NULL)
				pfast=pfast->next;
			if(pfast!=NULL)
				pfast=pfast->next;
			if(pfast==NULL)
				return false;
			pslow = pslow->next;
		}while(pfast != pslow);
		return true;
    }
};