chore: 更新部分写法

Author: yi-ge

README.md

@@ -203,7 +203,7 @@ TypeScript / JavaScript 基础算法、数据结构练习，包含 LeetCode 或
     输入：nums = [6, 7, 0, 3, 4]， target = 5
     输出：false
 
-- [循环数组中的环-暂无解](src/array/circular-array-loop.ts)
+- [循环数组中的环](src/array/circular-array-loop.ts)
 
   - LeetCode 457. 环形数组循环 <https://leetcode-cn.com/problems/circular-array-loop/>
   - LintCode 1229. 循环数组中的环 <https://www.lintcode.com/problem/circular-array-loop/description>

src/list/delete-node-in-a-linked-list.ts

@@ -7,14 +7,14 @@ import ListNode from "../lib/ListNode"
 /**
  Do not return anything, modify it in-place instead.
  */
-export function deleteNode (root: ListNode | null): void {
-  if (root === null) {
+export function deleteNode (node: ListNode | null): void {
+  if (node === null) {
     return
   }
-  if (root.next === null) {
-    root = null
+  if (node.next === null) {
+    node = null
     return
   }
-  root.val = root.next.val
-  root.next = root.next.next
+  node.val = node.next.val
+  node.next = node.next.next
 }

src/list/linked-list-cycle.ts

@@ -1,21 +1,27 @@
 import ListNode from '../lib/ListNode'
 
-/**
- * @param {ListNode} head
- * @return {boolean}
- */
-export const hasCycle = function (head: ListNode | null): boolean {
-  if (!head || !head.next) return false
-
-  // 双指针解法
-  let slow: ListNode = head
-  let fast: ListNode = head.next
+// export const hasCycle = function (head: ListNode | null): boolean {
+//   const set = new Set()
+//   let current = head
+//   while (current && current.next !== null) {
+//     if (set.has(current)) {
+//       return true
+//     }
+//     set.add(current)
+//     current = current.next
+//   }
+//   return false
+// }
 
-  while (fast && fast.next && fast.next.next && slow!.next) {
-    if (fast.next === slow || fast === slow) return true
-    fast = fast!.next!.next
-    slow = slow!.next
+export const hasCycle = function (head: ListNode | null): boolean {
+  let slow = head
+  let fast = head?.next
+  while (slow?.next && fast?.next?.next) {
+    if (fast.next === slow || slow === fast) return true
+    slow = slow.next
+    fast = fast.next.next
   }
-
   return false
 }
+
+