[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;
    }
};

　　

　　

　　

相关题目：

[LeetCode] 158. Read N Characters Given Read4 II - Call multiple times