1656. Design an Ordered Stream

1656. 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

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 #

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);
 */

⬅️上一页

下一页➡️

Calendar Apr 17, 2021
Edit Edit this page
本站总访问量:  次 您是本站第  位访问者