leetcode 141. Linked List Cycle

Discription

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

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

My solution:None

• 检测链表中有无cycle,可以用DPS方法,对每个node做访问过标记{0,1,2},但这需要额外空间.

Discuss

大神采用的方式是两个指针,在整个链表图上跑,一个fast,一个slow.
注意到因为有环的存在,在遍历程中,必然会掉入循环. 最终fast和slow在圈里一直跑动,fast会将slow套圈,也就是再次相遇!!
这与初中做的数学题问何时套圈的思路一样,考虑的应该是:场景设定为环,有怎样的区别于直线的特征?
很显然就是旧地重游,可以用不同速度区分
总之,这个转换的思路很巧妙!

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */ /** use faster and lower runner solution. (2 pointers) the faster one move 2 steps, and slower one move only one step. if there's a circle, the faster one will finally "catch" the slower one.  (the distance between these 2 pointers will decrease one every time.) if there's no circle, the faster runner will reach the end of linked list. (NULL) */class Solution {public:    bool hasCycle(ListNode *head) {        if(head == NULL || head -> next == NULL)                return false;        ListNode *fast = head;        ListNode *slow = head;        while(fast -> next && fast -> next -> next){            fast = fast -> next -> next;            slow = slow -> next;            if(fast == slow)                return true;        }        return false;    }};

Reference

• leetcode 141