add: 0141 solution js edition

Author: royIdoodle

docs/Leetcode_Solutions/JavaScript/0141._Linked_List_Cycle.md

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+# 001. Two Sum
+
+**<font color=red>难度: Easy</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/linked-list-cycle/
+
+> 内容描述
+
+```
+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.
+```
+
+![img](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png)
+
+**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.
+```
+
+![img](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png)
+
+**Example 3:**
+
+```
+Input: head = [1], pos = -1
+Output: false
+Explanation: There is no cycle in the linked list.
+```
+
+![img](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png)
+
+ 
+
+**Follow up:**
+
+Can you solve it using *O(1)* (i.e. constant) memory?
+
+
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
+
+使用`快慢指针`的思路进行解题。就像两个运动员在同一个环形赛道上赛跑，如果一个运动员跑的快，一个跑得慢，最后两个运动员一定会相遇。
+
+下面代码中的`fast`每次会走两步，而`slow`每次会走一步，如果`fast`没有`next`节点，自然没有环；如果`fast`等于`slow`说明二者相遇，最终为表明存在环。
+
+
+
+#### 执行结果
+
+执行用时 :**92 ms**, 在所有 JavaScript 提交中击败了94.16%的用户
+
+内存消耗 :**36.6 MB**, 在所有 JavaScript 提交中击败了51.93%
+
+
+
+代码：
+
+```javascript
+/**
+ * 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 === null || head.next === null) {
+    return false
+  }
+  
+  let slow = head
+  let fast = head.next
+  
+  while (slow !== fast) {
+    if (fast === null || fast.next === null) {
+      return false
+    }
+    slow = slow.next
+    fast = fast.next.next
+  }
+  return true
+};
+```
+