Given a file and assume that you can only read the file using a given method read4
, implement a method read
to read n characters. Your method read
may be called multiple times.
Method read4:
The API read4
reads 4 consecutive characters from the file, then writes those characters into the buffer array buf
.
The return value is the number of actual characters read.
Note that read4()
has its own file pointer, much like FILE *fp
in C.
Definition of read4:
Parameter: char[] buf4 Returns: int Note: buf4[] is destination not source, the results from read4 will be copied to buf4[]
Below is a high level example of how read4
works:
File file("abcde"); // File is "abcde", initially file pointer (fp) points to 'a'
char[] buf = new char[4]; // Create buffer with enough space to store characters
read4(buf4); // read4 returns 4. Now buf = "abcd", fp points to 'e'
read4(buf4); // read4 returns 1. Now buf = "e", fp points to end of file
read4(buf4); // read4 returns 0. Now buf = "", fp points to end of file
Method read:
By using the read4
method, implement the method read
that reads n characters from the file and store it in the buffer array buf
. Consider that you cannot manipulate the file directly.
The return value is the number of actual characters read.
Definition of read:
Parameters: char[] buf, int n Returns: int Note: buf[] is destination not source, you will need to write the results to buf[]
Example 1:
File file("abc"); Solution sol; // Assume buf is allocated and guaranteed to have enough space for storing all characters from the file. sol.read(buf, 1); // After calling your read method, buf should contain "a". We read a total of 1 character from the file, so return 1. sol.read(buf, 2); // Now buf should contain "bc". We read a total of 2 characters from the file, so return 2. sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
Example 2:
File file("abc"); Solution sol; sol.read(buf, 4); // After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3. sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
Note:
- Consider that you cannot manipulate the file directly, the file is only accesible for
read4
but not forread
. - The
read
function may be called multiple times. - Please remember to RESET your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see here for more details.
- You may assume the destination buffer array,
buf
, is guaranteed to have enough space for storing n characters. - It is guaranteed that in a given test case the same buffer
buf
is called byread
.
A:
这个题目tricky 的地方在于: 以前遗留下来的 buffer ,可能这次还读取不完。
/** * The read4 API is defined in the parent class Reader4. * int read4(char *buf4); */ class Solution { public: /** * @param buf Destination buffer * @param n Number of characters to read * @return The number of actual characters read */ int read(char *buf, int n) { int count = 0; while(count < n){ // first read up the values in buf4 for(; iStart<preCount && count < n;iStart++){ *(buf+count++) = *(buf4+iStart); } if(count>=n){ break; } // read another round of buf if(iStart >= preCount){ preCount = read4(buf4); iStart = 0; } if(preCount==0) break; // if no data readin anymore } return count; } private: char * buf4 = new char[4]; int preCount = 0; int iStart = 0; // position that we can start to read new char };错误:
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