【LeetCode】Linked List Cycle

最新推荐文章于 2021-11-01 14:38:45 发布

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本文介绍了使用快慢指针法来判断链表是否存在循环，并提供了示例代码和测试用例，详细解释了算法原理及注意事项。

参考链接：

http://blog.youkuaiyun.com/lidalong0408/article/details/14103295

http://blog.youkuaiyun.com/sysucph/article/details/15378043

题目描述

Linked List Cycle

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

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

题目分析

判断链表是否有环，可使用快慢指针，如果有环，快指针一定可以追上慢指针。

总结

这个题在写是一个地方出错了，就是开始把快慢指针的初始值设为haed，进行while循环后先判断slow==fast，所以一定会返回。

就改为先移动指针再判断

		while(fast->next != NULL && fast->next->next != NULL) 
		{
			slow = slow->next;
	  		fast = fast->next;
	  		fast = fast->next; 
			if(fast == slow)			//一开始都是head，所以不能直接判断，需要进行一次移动后再判断 
			  return true;	  		
		}

示例代码

/*
编译环境CFree 5.0 

*/
#include 
   
    
#include 
    
     

using namespace std;


struct ListNode{
	int val;
	ListNode *next;
	ListNode(int x):val(x),next(NULL){} 
} ;


void printList(ListNode *head,char *name)
{
	printf("%s\n",name);
	while(head != NULL)
	{
		printf("%3d,",head->val);
		head = head->next;
	}
	printf("\n");
}


class Solution { 
public: 
    bool hasCycle(ListNode *head) {
		if(head == NULL)
			return false;
		ListNode *slow = head;
		ListNode *fast = head;
		while(fast->next != NULL && fast->next->next != NULL) 
		{
			slow = slow->next;
	  		fast = fast->next;
	  		fast = fast->next; 
			if(fast == slow)			//一开始都是head，所以不能直接判断，需要进行一次移动后再判断 
			  return true;	  		
		}
		return false; 
    }
}; 

ListNode* listCreate(vector
     
       &list, int arr[], int n)
{
	if(n <= 0)
		return NULL;
		
	for(int i=0;i
      
       next = &list[i];
		head = head->next;
	}
	head = &list[0];
	return head;
}

//有环 
void test0()
{
	vector
       
         list;
	int arr[10] = {0,1,2,3,4,5,6,7,8,9};
	ListNode *head = listCreate(list,arr, 10);
	printList(head,"head");
	list[9].next = &list[3];//制造一个环 
	Solution so;
	if(so.hasCycle(head) == false)
		printf("------------------------failed\n");
	else
		printf("------------------------passed\n");

}
//无环 
void test1()
{
	vector
        
          list; int arr[10] = {0,1,2,3,4,5,6,7,8,9}; ListNode *head = listCreate(list,arr, 10); printList(head,"head"); //list[9].next = &list[3];//制造一个环 Solution so; if(so.hasCycle(head) == true) printf("------------------------failed\n"); else printf("------------------------passed\n"); } int main() { test0(); test1(); return 0; }