leetcode #141 in cpp

xiaolienahu

于 2016-06-25 12:01:24 发布

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分类专栏： interview 文章标签： cpp leetcode

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本文介绍了一种使用快慢指针的方法来判断链表中是否存在循环。通过一个移动速度为一步的慢指针和一个移动速度为两步的快指针，在遍历过程中如果存在循环，两个指针最终会相遇；若不存在循环，则快指针会首先到达链表尾部。

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

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

Solution;

Make a slow pointer which step 1 in each iteration and a fast pointer which step 2 in each iteration.

If there is a cycle, slow would meet fast at some point.

Otherwise fast would reach NULL

Code:

/**
 * 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) {
        ListNode *fast = head; 
        ListNode *slow = head;
        while(fast && fast->next){
            slow = slow->next; 
            fast = fast->next-> next;
            if(slow == fast) return true; 
            
        }
        return false;
        
    }
};