【LeetCode 0141】【链表】【双指针之快慢指针】判断给定单链表是否存在环

141. Linked List Cycle

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

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to. Note that pos is not passed as a parameter.

Return true* if there is a cycle in the linked list*. Otherwise, return false.

Example 1:

**Input:** head = [3,2,0,-4], pos = 1
**Output:** true
**Explanation:** There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

Example 2:

**Input:** head = [1,2], pos = 0
**Output:** true
**Explanation:** There is a cycle in the linked list, where the tail connects to the 0th node.

Example 3:

**Input:** head = [1], pos = -1
**Output:** false
**Explanation:** There is no cycle in the linked list.

Constraints:

• The number of the nodes in the list is in the range [0, 10^4].
• -10^5 <= Node.val <= 10^5
• pos is -1 or a valid index in the linked-list.

Follow up: Can you solve it using O(1) (i.e. constant) memory?

Idea

* 如果链表不存在环，则迭代会遍历终止（末尾节点为null）
* 如果存在环，则有且只有一个环，而且只能是在链表尾部形成环
  类似龟兔赛跑，慢指针迟早会赶上快指针，如果存在快指针于慢指针重叠，则说明存在环。

JavaScript Solution

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
    if(!head){
        return false
    }
    let [slow,fast] = [head,head.next]
    while(fast && fast.next){
        if(slow==fast){
            return true
        }
        slow = slow.next
        fast = fast.next.next
    }
    return false
};