158. Read N Characters Given read4 II - Call Multiple Times
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 four consecutive characters from file
, then writes those characters into the buffer array buf4
.
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
buf4[] is a destination, not a 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[] buf4 = new char[4]; // Create buffer with enough space to store characters
read4(buf4); // read4 returns 4. Now buf4 = "abcd", fp points to 'e'
read4(buf4); // read4 returns 1. Now buf4 = "e", fp points to end of file
read4(buf4); // read4 returns 0. Now buf4 = "", fp points to end of file
Method read:
By using the read4
method, implement the method read that reads n
characters from file
and store it in the buffer array buf
. Consider that you cannot manipulate file
directly.
The return value is the number of actual characters read.
Definition of read:
Parameters: char[] buf, int n
Returns: int
buf[] is a destination, not a source. You will need to write the results to buf[].
Note:
- Consider that you cannot manipulate the file directly. The file is only accessible 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 storingn
characters. - It is guaranteed that in a given test case the same buffer
buf
is called byread
.
Example 1:
Input: file = "abc", queries = [1,2,1]
Output: [1,2,0]
Explanation: The test case represents the following scenario:
File file("abc");
Solution sol;
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.
Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.
Example 2:
Input: file = "abc", queries = [4,1]
Output: [3,0]
Explanation: The test case represents the following scenario:
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.
Constraints:
1 <= file.length <= 500
file
consist of English letters and digits.1 <= queries.length <= 10
1 <= queries[i] <= 500
Solution:
/**
* The read4 API is defined in the parent class Reader4.
* int read4(char[] buf4);
*/
/*
[a, b, c, d, e]
buff
0 1 2 3
[ , , , ]
*/
public class Solution extends Reader4 {
/**
* @param buf Destination buffer
* @param n Number of characters to read
* @return The number of actual characters read
*/
private int buffPtr = 0;
// This is a pointer to keep track of the current position in the `buff` array.
// It helps to know which character in `buffer` should be read next;
private int buffCnt = 0;
// This keeps track of the number of characters actually returned by the `read4`
// method. It might be less than 4 if the end of the file is reached.
private char[] buff = new char[4];
// This is a temporary buffer of size 4 that stores characters read from the `read4`
// method
public int read(char[] buf, int n) { // " a b c"
int ptr = 0; // --> number of char we read
while (ptr < n){
if (buffPtr == 0){
buffCnt = read4(buff); // it will update the buff and return the point
}
if (buffCnt == 0){
break; // nothing to read
}
while(ptr < n && buffPtr < buffCnt){
buf[ptr] = buff[buffPtr];
ptr++;
buffPtr++;
}
if (buffPtr >= buffCnt){
buffPtr = 0;
}
}
return ptr;
}
}
做过类似
嗯...
又是看不懂题系列
158 是 157 的follow up