commit ddf756f53d24748551a00fba43bdbd1bb10c70d1
Author: mol <yunying_mol@163.com>
Date:   Mon Jul 31 13:57:28 2023 +0800

    重排链表

diff --git a/1.重排链表/复习.js b/1.重排链表/复习.js
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+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {void} Do not return anything, modify head in-place instead.
+ */
+var reorderList = function (head) {
+  let slow = head
+  let fast = head
+
+  // 使用快慢指针寻找链表中点
+  while (fast.next && fast.next.next) {
+    fast = fast.next.next
+    slow = slow.next
+  }
+
+  // 拆分成前后链表
+  let cur = slow.next
+  slow.next = null
+
+  // 反转后链表
+  let pre = null
+  while (cur) {
+    let next = cur.next // 4 null
+    cur.next = pre // null 3
+    pre = cur // 3 4
+    cur = next // 4 null
+    // null <- 3 <- 4
+  }
+
+  // 合并前后链表
+  let lp = head
+  while (lp && pre) {
+    let ln = lp.next
+    let rn = pre.next
+    lp.next = pre
+    pre.next = ln
+    lp = ln
+    pre = rn
+  }
+
+  return head
+};
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diff --git a/1.重排链表/笔记.md b/1.重排链表/笔记.md
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+## 快慢指针
+快慢指针中的快慢指的是移动的步长，即每次向前移动速度的快慢。例如可以让快指针每次沿链表向前移动2，慢指针每次向前移动1次。
+
+### 快慢指针的应用
+#### 判断单链表是否为循环链表
+让快慢指针从链表头开始遍历，快指针向前移动两个位置，慢指针向前移动一个位置;如果快指针到达NULL，说明链表以NULL为结尾，不是循环链表。如果 快指针追上慢指针，则表示出现了循环。
+```
+fast=slow=head;
+fast=fast->next->next;
+slow=slow->next;
+while（true）{
+if (fast==NULL || fast->next==NULL)
+return false;
+else if (fast==slow || fast->next==slow)
+return true;
+else{
+fast=fast->next->next;
+slow=slow->next;
+}
+}
+```
+#### 在有序链表中寻找中位数
+该方法在不借助计数器变量实现寻找中位数的功能。原理是：快指针的移动速度是慢指针移动速度的2倍，因此当快指针到达链表尾时，慢指针到达中点。程序还要考虑链表结点个数的奇偶数因素，当快指针移动x次后到达表尾（1+2x），说明链表有奇数个结点，直接返回慢指针指向的数据即可。如果快指针是倒数第二个结点，说明链表结点个数是偶数，这时可以根据“规则”返回上中位数或下中位数或（上中位数+下中位数）的一半。
+```
+while (fast&&slow)
+{
+if (fast->next==NULL)
+return slow ->data;
+else if (fast->next!= NULL && fast->next->next== NULL)
+return (slow ->data + slow ->next->data)/2;
+else
+{
+fast= fast->next;
+fast= fast->next;
+slow = slow ->next;
+}
+}
+```
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diff --git a/1.重排链表/解1.js b/1.重排链表/解1.js
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+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {void} Do not return anything, modify head in-place instead.
+ */
+var reorderList = function (head) {
+  const node = head
+  let lastNode = getLastNode(node)
+  let currentNode = node
+  while (!(currentNode === lastNode || currentNode.next === lastNode)) {
+    const secNode = currentNode.next
+    const lastPrevNode = getPrevNode(node, lastNode)
+    currentNode.next = lastNode
+    lastNode.next = secNode
+    currentNode = secNode
+    lastNode = lastPrevNode
+    lastNode.next = null
+  }
+  return head
+};
+
+function getLastNode(node) {
+  let lastNode = node;
+  while (lastNode.next) {
+    lastNode = lastNode.next
+  }
+  return lastNode
+}
+
+function getPrevNode(head, node) {
+  let currentNode = head;
+  while (currentNode && currentNode.next !== node) {
+    currentNode = currentNode.next
+  }
+  return currentNode
+}
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diff --git a/1.重排链表/题目.md b/1.重排链表/题目.md
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+给定一个单链表 L 的头节点 head ，单链表 L 表示为：
+
+```
+L0 → L1 → … → Ln - 1 → Ln
+```
+请将其重新排列后变为：
+
+```
+L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …
+```
+不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
+
+ 
+
+示例 1：
+
+![1626420311-PkUiGI-image[1].png](https://picbed.hiiragi.club:8081/i/2023/07/31/64c74c89a9ff8.png)
+
+```
+输入：head = [1,2,3,4]
+输出：[1,4,2,3]
+```
+示例 2：
+
+![1626420320-YUiulT-image[1].png](https://picbed.hiiragi.club:8081/i/2023/07/31/64c74c8a1d9a8.png)
+
+```
+输入：head = [1,2,3,4,5]
+输出：[1,5,2,4,3]
+```
+ 
+
+提示：
+
+- 链表的长度范围为 `[1, 5 * 104]`
+
+- `1 <= node.val <= 1000`
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diff --git a/1.重排链表/题解.md b/1.重排链表/题解.md
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+方法一：快慢指针 + 反转链表 + 合并链表
+
+> 作者：lcbin
+链接：https://leetcode.cn/problems/reorder-list/solution/python3javacgo-yi-ti-yi-jie-kuai-man-zhi-t9u2/
+来源：力扣（LeetCode）
+著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
+我们先用快慢指针找到链表的中点，然后将链表的后半部分反转，最后将左右两个链表合并。
+
+```js
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {void} Do not return anything, modify head in-place instead.
+ */
+var reorderList = function (head) {
+    // 快慢指针找到链表中点
+    let slow = head;
+    let fast = head;
+    while (fast.next && fast.next.next) {
+        slow = slow.next;
+        fast = fast.next.next;
+    }
+
+    // cur 指向右半部分链表
+    let cur = slow.next;
+    slow.next = null;
+
+    // 反转右半部分链表
+    let pre = null;
+    while (cur) {
+        const t = cur.next;
+        cur.next = pre;
+        pre = cur;
+        cur = t;
+    }
+    cur = head;
+
+    // 此时 cur, pre 分别指向链表左右两半的第一个节点
+    // 合并
+    while (pre) {
+        const t = pre.next;
+        pre.next = cur.next;
+        cur.next = pre;
+        cur = pre.next;
+        pre = t;
+    }
+};
+```
+时间复杂度O(n)，其中 n 是链表的长度。空间复杂度O(1)
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