[LeetCode] 157. Read N Characters Given Read4 用Read4来读取N个字符

The API: int read4(char *buf) reads 4 characters at a time from a file.
The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.
By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file.
Note:
The read function will only be called once for each test case.

给一个Read4函数，每次可以从一个文件中最多读出4个字符，如果文件中的字符不足4个，返回当前剩余的字符。实现一个读取n个字符的函数。

用一个临时数组，存放每次read4读到字符，再用一个指针标记buf数组目前存储到的位置，然后将这个临时数组的内容存到buf相应的位置就行了。

需要注意两个corner case：

如果本次读到多个字符，但是我们只需要其中一部分就能完成读取任务时，我们要拷贝的长度是本次读到的个数和剩余所需个数中较小的

如果read4没有读满4个，说明数据已经读完，这时候对于读到的数据长度，因为也可能存在我们只需要其中一部分的情况，所以要返回总所需长度和目前已经读到的长度的较小的

Java: Time: O(n), Space: O(1)

public class Solution extends Reader4 {

    public int read(char[] buf, int n) {

        for(int i = 0; i < n; i += 4){

            char[] tmp = new char[4];

            // 将数据读入临时数组

            int len = read4(tmp);

            // 将临时数组拷贝至buf数组，这里拷贝的长度是本次读到的个数和剩余所需个数中较小的

            System.arraycopy(tmp, 0, buf, i, Math.min(len, n - i));

            // 如果读不满4个，说明已经读完了，返回总所需长度和目前已经读到的长度的较小的

            if(len < 4) return Math.min(i + len, n);

        }

        // 如果循环内没有返回，说明读取的字符是4的倍数

        return n;

    }

}

Python:

# The read4 API is already defined for you.

# @param buf, a list of characters

# @return an integer

def read4(buf):

    global file_content

    i = 0

    while i < len(file_content) and i < 4:

        buf[i] = file_content[i]

        i += 1

    if len(file_content) > 4:

        file_content = file_content[4:]

    else:

        file_content = ""

    return i

class Solution(object):

    def read(self, buf, n):

        """

        :type buf: Destination buffer (List[str])

        :type n: Maximum number of characters to read (int)

        :rtype: The number of characters read (int)

        """

        read_bytes = 0

        buffer = [''] * 4

        for i in xrange(n / 4 + 1):

            size = read4(buffer)

            if size:

                size = min(size, n-read_bytes)

                buf[read_bytes:read_bytes+size] = buffer[:size]

                read_bytes += size

            else:

                break

        return read_bytes

if __name__ == "__main__":

    global file_content

    buf = ['' for _ in xrange(100)]

    file_content = "a"

    print(buf[:Solution().read(buf, 9)])

    file_content = "abcdefghijklmnop"

    print(buf[:Solution().read(buf, 9)])

C++:

// Time:  O(n)

// Space: O(1)

int read4(char *buf);

class Solution {

public:

    /**

     * @param buf Destination buffer

     * @param n   Maximum number of characters to read

     * @return    The number of characters read

     */

    int read(char *buf, int n) {

        int read_bytes = 0;

        char buffer[4];

        for (int i = 0; i <= n / 4; ++i) {

            if (int size = read4(buffer)) {

                size = min(size, n - read_bytes);

                for (int j = 0; j < size; ++j) {

                    buf[read_bytes++] = buffer[j];

                }

            } else {

                break;

            }

        }

        return read_bytes;

    }

};

巴特西

[LeetCode] 157. Read N Characters Given Read4 用Read4来读取N个字符

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