【leetcode】1656. Design an Ordered Stream

题目如下：

There is a stream of n (idKey, value) pairs arriving in an arbitrary order, where idKey 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 idKey, String value) Inserts the pair (idKey, value) into the stream, then returns the largest possible chunk of currently inserted values that appear next in the order.

 

Example:

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.

解题思路：本题最难的地方在于能不能读懂题目，英文不太好理解的话可以看中文版的。

代码如下：

class OrderedStream(object):

    def __init__(self, n):
        """
        :type n: int
        """
        self.ptr = 1
        self.l = [None] * (n + 1)

    def insert(self, id, value):
        """
        :type id: int
        :type value: str
        :rtype: List[str]
        """
        res = []
        self.l[id] = value
        start = self.ptr
        for i in range(start,len(self.l)):
            if i == start and self.l[i] == None:
                return []
            elif i == start and self.l[i] != None:
                res.append(self.l[i])
            elif self.l[i] == None:
                self.ptr = i
                break
            else:
                res.append(self.l[i])
                self.ptr = i+1
        return res
        


# Your OrderedStream object will be instantiated and called as such:
# obj = OrderedStream(n)
# param_1 = obj.insert(id,value)