1. 问题

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:
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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.

2. 解题思路

方法

1.定义空的string数组为res，指针为ptr
2.指针起始位置为0，res的数组长度等于n
3. 新建一个list
4. res的长度为idKey-1，字符串数据插入到第 K 个位置（idKey 减 1）
5.如果位置已满，指针一直向右移动,在list添加res索引位置添加相应的值
6…返回list

3. 代码

代码：

class OrderedStream {
    int ptr;//指针起始位置
    String[] res;//1.定义空的string数组为res，指针为ptr
    
    public OrderedStream(int n) {
        ptr = 0;//2.指针起始位置为0，res的数组长度等于n
        res = new String[n];
    }
    
    public List<String> insert(int idKey, String value) {
        List<String> list = new ArrayList<>();//3. 新建一个list
        
        res[idKey - 1] = value; //4. res的长度为idKey-1，字符串数据插入到第 K 个位置（idKey 减 1）
        while (ptr < res.length && res[ptr] != null) {//5.如果位置已满，请尝试将指针一直向右移动,在list添加res索引位置对应的值
            list.add(res[ptr]);
            ptr++;
        }
        return list;//6.返回list
    }
}

/**
 * Your OrderedStream object will be instantiated and called as such:
 * OrderedStream obj = new OrderedStream(n);
 * List<String> param_1 = obj.insert(idKey,value);
 */