[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)

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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:

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# 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++:

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// 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