141. Linked List Cycle（Easy）2019.7.10
题目地址https://leetcode.com/problems/linked-list-cycle/
Given a linked list, determine if it has a cycle in it.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it using O(1) (i.e. constant) memory?
solution
解法一
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode p = head; //游标指针
while (p != null && p.next != null) //注意循环的终止条件！！！！
{
p.val = Integer.MIN_VALUE; //将结点里面的值设为一个超小的值
if (p.next.val == Integer.MIN_VALUE) //判断是否有环
return true;
else
p = p.next;
}
return false;
}
}
解法二
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode fast = head, slow = head; //设置快慢指针，fast和slow
boolean flag = false;
while (slow != null && slow.next != null && fast != null && fast.next != null) //要格外注意循环的终止条件！！！！！
{
if (slow == fast && flag == true) //判断两指针是否相遇，且不是在初次出发处
return true;
else
{
flag = true;
slow = slow.next;
fast = fast.next.next;
}
}
return false;
}
}
reference
https://leetcode.com/problems/linked-list-cycle/solution/
总结
题意是给定一个链表，判断此链表里面是否有环，题目的附加要求是只用O(1)的空间解决此题。
- 解法一是直接遍历链表，将链表里面当前结点的val值设置为一个极小的数Integer.MIN_VALUE，然后判断当前结点的下一个结点值val是否为Integer.MIN_VALUE,若是，则返回true，否则再判断下一个结点，如果循环结束还没找到，则返回false。
- 解法二利用了快慢指针有环必相遇的思想。首先设置一个slow指针，每次走一步，然后设置一个fast指针，每次走两步，若链表有环，则若干步后两个指针必相遇。解法里面设置了一个flag变量来控制第一次slow==fast时不会返回true。此种解法要格外注意while循环的结束条件，稍不留意，就会出错。
Notes
1.链表问题要格外注意循环的终止条件；