leetcode update

content/posts/leetcode.md

@@ -1861,3 +1861,114 @@ func reverse(head *ListNode) *ListNode{
     return prev
 }
 ```
+
+## day26 2024-03-23
+
+### 143. Reorder
+
+You are given the head of a singly linked-list. The list can be represented as:
+
+> L0 → L1 → … → Ln - 1 → Ln
+
+Reorder the list to be on the following form:
+
+> L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …
+
+You may not modify the values in the list's nodes. Only nodes themselves may be changed.
+
+![0323xGTn0liJdHDU](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0323xGTn0liJdHDU.png)
+
+### 题解
+
+拿到本题, 首先想到的是将链表中的全部数据保存到数组中, 然后通过同时从前后遍历数组并且给原来的链表赋值的方式, 即可快速解决本题, 如果不使用额外的空间, 可以使用快慢指针, 找到链表的中间节点, 从中间节点开始将链表的后半部分反向, 用两个指针分别从前半部分的开始和后半部分的末尾开始遍历, 并构造新链表即可. 也可以构造一个栈, 将链表的后半部分放入栈中, 然后从顶端一边出栈一边构造新链表即可. 题目中要求不能直接修改节点的值, 因此采用第三种思路.
+
+### 代码
+
+```go
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func reorderList(head *ListNode)  {
+    slow := head
+    fast := head
+    if head.Next != nil{
+        fast = head.Next
+    }
+    stack := []*ListNode{}
+    for fast != nil && fast.Next != nil{
+        fast = fast.Next.Next
+        slow = slow.Next
+    }
+    odd := false
+    if fast  == nil{
+        odd = true
+    }
+    // construct pointer stack
+    slow = slow.Next
+    for slow != nil{
+        stack = append(stack, slow)
+        slow = slow.Next
+    }
+    slow = head
+    temp := head.Next
+    temp2 := head
+    flag := 1
+    for i:=len(stack)-1;i>=0;i--{
+        if flag == 1{
+            stack[i].Next = nil
+            slow.Next = stack[i]
+            slow = slow.Next
+            flag = 0
+        }else{
+            temp2 = temp.Next
+            temp.Next = nil
+            slow.Next = temp
+            slow = slow.Next
+            flag = 1
+            i++
+            temp = temp2
+        }
+    }
+    if odd{
+        temp.Next = nil
+        slow.Next = temp
+    }
+    return
+}
+
+
+```
+
+### 总结
+
+查看最快速度代码, 使用了一个数组先将链表的全部节点指针留一个间隔存放在数组中, 然后再逆序遍历数组将后半节点的指针逆序插入前面留出的空格中, 最后从头遍历数组, 连接整个链表即可. 这样写得出的代码比较简洁.
+
+```go
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func reorderList(head *ListNode) {
+	list := []*ListNode{}
+
+	for node := head; node != nil; node = node.Next {
+		list = append(list, node)
+		list = append(list, nil)
+	}
+
+	for i := 1; i < len(list) - i - 1; i += 2 {
+		list[i] = list[len(list) - i - 1]
+	}
+
+	for i := 0; list[i] != nil; i++ {
+		list[i].Next = list[i + 1]
+	}
+}
+```