题目如下:
There is a stream of
n
(idKey, value)
pairs arriving in an arbitrary order, whereidKey
is an integer between1
andn
andvalue
is a string. No two pairs have the sameid
.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 taken
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 uniqueid.
- Exactly
n
calls will be made toinsert
.
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)