問題：

Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

解法1：

使用兩個步伐不一樣的指標遍歷連結串列（eg：a每步走一格，b每步走兩格），如果連結串列存在環路，兩個指標總會相遇

/**
 * 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) {
        if(head == nullptr)
            return false;
        ListNode* walker = head;
        ListNode* runner = head;
        while(runner->next != nullptr && runner->next->next != nullptr) {
            walker = walker->next;
            runner = runner->next->next;
            if(walker==runner) return true;
        }
        return false;
    }
};
 

解法2：

過河拆橋。每走過一個節點，就將該節點的next指標置為head（相當於將已經走過的節點做標記），如果遇到某個節點的next為head，則說明之前走到過這個節點，有環路。

public class Solution {
    public boolean hasCycle(ListNode head) {
    	ListNode p = head,pre = head;
    	while(p != null && p.next != null){
    		if (p.next == head) return true;
    		p = p.next;
    		pre.next = head;
    		pre = p;
    	}
        return false;
    }
}