Add solution 1656、1657

leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream.go

@@ -0,0 +1,37 @@
+package leetcode
+
+type OrderedStream struct {
+	ptr    int
+	stream []string
+}
+
+func Constructor(n int) OrderedStream {
+	ptr, stream := 1, make([]string, n+1)
+	return OrderedStream{ptr: ptr, stream: stream}
+}
+
+func (this *OrderedStream) Insert(id int, value string) []string {
+	this.stream[id] = value
+	res := []string{}
+	if this.ptr == id || this.stream[this.ptr] != "" {
+		res = append(res, this.stream[this.ptr])
+		for i := id + 1; i < len(this.stream); i++ {
+			if this.stream[i] != "" {
+				res = append(res, this.stream[i])
+			} else {
+				this.ptr = i
+				return res
+			}
+		}
+	}
+	if len(res) > 0 {
+		return res
+	}
+	return []string{}
+}
+
+/**
+ * Your OrderedStream object will be instantiated and called as such:
+ * obj := Constructor(n);
+ * param_1 := obj.Insert(id,value);
+ */

leetcode/1656.Design-an-Ordered-Stream/1656. Design an Ordered Stream_test.go

@@ -0,0 +1,21 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem1656(t *testing.T) {
+	obj := Constructor(5)
+	fmt.Printf("obj = %v\n", obj)
+	param1 := obj.Insert(3, "ccccc")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(1, "aaaaa")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(2, "bbbbb")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(5, "eeeee")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+	param1 = obj.Insert(4, "ddddd")
+	fmt.Printf("param_1 = %v obj = %v\n", param1, obj)
+}

leetcode/1656.Design-an-Ordered-Stream/README.md

@@ -0,0 +1,106 @@
+# [1656. Design an Ordered Stream](https://leetcode.com/problems/design-an-ordered-stream/)
+
+## 题目
+
+There is a stream of `n` `(id, value)` pairs arriving in an **arbitrary** order, where `id` is an integer between `1` and `n` and `value` is a string. No two pairs have the same `id`.
+
+Design a stream that returns the values in **increasing order of their IDs** by returning a **chunk** (list) of values after each insertion. The concatenation of all the **chunks** should result in a list of the sorted values.
+
+Implement the `OrderedStream` class:
+
+- `OrderedStream(int n)` Constructs the stream to take `n` values.
+- `String[] insert(int id, String value)` Inserts the pair `(id, value)` into the stream, then returns the **largest possible chunk** of currently inserted values that appear next in the order.
+
+**Example:**
+
+![https://assets.leetcode.com/uploads/2020/11/10/q1.gif](https://assets.leetcode.com/uploads/2020/11/10/q1.gif)
+
+```
+Input
+["OrderedStream", "insert", "insert", "insert", "insert", "insert"]
+[[5], [3, "ccccc"], [1, "aaaaa"], [2, "bbbbb"], [5, "eeeee"], [4, "ddddd"]]
+Output
+[null, [], ["aaaaa"], ["bbbbb", "ccccc"], [], ["ddddd", "eeeee"]]
+
+Explanation
+// Note that the values ordered by ID is ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"].
+OrderedStream os = new OrderedStream(5);
+os.insert(3, "ccccc"); // Inserts (3, "ccccc"), returns [].
+os.insert(1, "aaaaa"); // Inserts (1, "aaaaa"), returns ["aaaaa"].
+os.insert(2, "bbbbb"); // Inserts (2, "bbbbb"), returns ["bbbbb", "ccccc"].
+os.insert(5, "eeeee"); // Inserts (5, "eeeee"), returns [].
+os.insert(4, "ddddd"); // Inserts (4, "ddddd"), returns ["ddddd", "eeeee"].
+// Concatentating all the chunks returned:
+// [] + ["aaaaa"] + ["bbbbb", "ccccc"] + [] + ["ddddd", "eeeee"] = ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"]
+// The resulting order is the same as the order above.
+
+```
+
+**Constraints:**
+
+- `1 <= n <= 1000`
+- `1 <= id <= n`
+- `value.length == 5`
+- `value` consists only of lowercase letters.
+- Each call to `insert` will have a unique `id.`
+- Exactly `n` calls will be made to `insert`.
+
+## 题目大意
+
+有 n 个 (id, value) 对，其中 id 是 1 到 n 之间的一个整数，value 是一个字符串。不存在 id 相同的两个 (id, value) 对。
+
+设计一个流，以 任意 顺序获取 n 个 (id, value) 对，并在多次调用时 按 id 递增的顺序 返回一些值。
+
+实现 OrderedStream 类：
+
+- OrderedStream(int n) 构造一个能接收 n 个值的流，并将当前指针 ptr 设为 1 。
+- String[] insert(int id, String value) 向流中存储新的 (id, value) 对。存储后：
+如果流存储有 id = ptr 的 (id, value) 对，则找出从 id = ptr 开始的 最长 id 连续递增序列 ，并 按顺序 返回与这些 id 关联的值的列表。然后，将 ptr 更新为最后那个 id + 1 。
+否则，返回一个空列表。
+
+## 解题思路
+
+- 设计一个具有插入操作的 Ordered Stream。insert 操作先在指定位置插入 value，然后返回当前指针 ptr 到最近一个空位置的最长连续递增字符串。如果字符串不为空，ptr 移动到非空 value 的后一个下标位置处。
+- 简单题。按照题目描述模拟即可。注意控制好 ptr 的位置。
+
+## 代码
+
+```go
+package leetcode
+
+type OrderedStream struct {
+	ptr    int
+	stream []string
+}
+
+func Constructor(n int) OrderedStream {
+	ptr, stream := 1, make([]string, n+1)
+	return OrderedStream{ptr: ptr, stream: stream}
+}
+
+func (this *OrderedStream) Insert(id int, value string) []string {
+	this.stream[id] = value
+	res := []string{}
+	if this.ptr == id || this.stream[this.ptr] != "" {
+		res = append(res, this.stream[this.ptr])
+		for i := id + 1; i < len(this.stream); i++ {
+			if this.stream[i] != "" {
+				res = append(res, this.stream[i])
+			} else {
+				this.ptr = i
+				return res
+			}
+		}
+	}
+	if len(res) > 0 {
+		return res
+	}
+	return []string{}
+}
+
+/**
+ * Your OrderedStream object will be instantiated and called as such:
+ * obj := Constructor(n);
+ * param_1 := obj.Insert(id,value);
+ */
+```