LeetCode Solutions
157. Read N Characters Given Read4
Time: $O(n)$ Space: $O(1)$
/**
* 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) {
char* buf4 = new char[4];
int i4 = 0; // buf4's index
int n4 = 0; // buf4's size
int i = 0; // buf's index
while (i < n) {
if (i4 == n4) { // All characters in buf4 are consumed
i4 = 0; // Reset buf4's index
n4 = read4(buf4); // Read 4 (or less) chars from file to buf4
if (n4 == 0) // Reach the EOF
return i;
}
buf[i++] = buf4[i4++];
}
return i;
}
};
/**
* The read4 API is defined in the parent class Reader4.
* int read4(char[] buf4);
*/
public class Solution extends Reader4 {
/**
* @param buf Destination buffer
* @param n Number of characters to read
* @return The number of actual characters read
*/
public int read(char[] buf, int n) {
char[] buf4 = new char[4];
int i4 = 0; // buf4's index
int n4 = 0; // buf4's size
int i = 0; // buf's index
while (i < n) {
if (i4 == n4) { // All characters in buf4 are consumed
i4 = 0; // Reset buf4's index
n4 = read4(buf4); // Read 4 (or less) chars from file to buf4
if (n4 == 0) // Reach the EOF
return i;
}
buf[i++] = buf4[i4++];
}
return i;
}
}
"""
The read4 API is already defined for you.
def read4(buf4: List[chr]) -> int:
# Below is an example of how the read4 API can be called.
file = File("abcdefghijk") # File is "abcdefghijk", initially file pointer (fp) points to 'a'
buf4 = [' '] * 4 # Create buffer with enough space to store characters
read4(buf4) # Read4 returns 4. Now buf = ['a','b','c','d'], fp points to 'e'
read4(buf4) # Read4 returns 4. Now buf = ['e','f','g','h'], fp points to 'i'
read4(buf4) # Read4 returns 3. Now buf = ['i','j','k',...], fp points to end of file
"""
class Solution:
def read(self, buf: List[str], n: int) -> int:
buf4 = [' '] * 4
i4 = 0 # buf4's index
n4 = 0 # buf4's size
i = 0 # buf's index
while i < n:
if i4 == n4: # All characters in buf4 are consumed
i4 = 0 # Reset buf4's index
n4 = read4(buf4) # Read 4 (or less) chars from file to buf4
if n4 == 0: # Reach the EOF
return i
buf[i] = buf4[i4]
i += 1
i4 += 1
return i