题目：
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.
链接： http://leetcode.com/problems/read-n-characters-given-read4/
题解：
英文不好，很难理解题意。简单讲就是已经有个API – read4(char[] buf)，每次从文件里读取最多4个bytes到char[] buf里。要求根据read4来实现read(char[] buf, n)，就是可以从文件中读取n个字符。这里我们要注意的就是，假如read4读取的字符数小于4，那么即到了文件尾部，我们可以使用一个变量EOF来记录这个时刻。除此之外就是一些判断和拷贝了。这里用到了System.arraycopy()。
Time Complexity – O(n)，Space Complexity – O(1)
/* The read4 API is defined in the parent class Reader4.
int read4(char[] buf); */public class Solution extends Reader4 {
/**
* @param buf Destination buffer
* @param n Maximum number of characters to read
* @return The number of characters read
*/
public int read(char[] buf, int n) {
char[] read4Buffer = new char[4];
boolean EOF = false;
int bytesRead = 0; while(!EOF && bytesRead < n) {
int read4Bytes = read4(read4Buffer);
if(read4Bytes < 4)
EOF = true;
int bytes = Math.min(n - bytesRead, read4Bytes);
System.arraycopy(read4Buffer, 0, buf, bytesRead, bytes);
bytesRead += bytes;
} return bytesRead;
}
}
二刷:
- 先建立一个read4Buf来保存使用read4 api之后读取的数据。 并且我们建立一个boolean EOF来代表是否读到了文件末尾，即read4返回的值小于4。
- 接下来我们用一个while循环来读取
- 当read4返回的值小于4的时候，我们设置EOF = true，即在下一次循环跳出
- 接下来我们来判断每次究竟要读取多少个bytes， 这个bytesToRead = Math.min(n – bytesRead，read4Bytes)， 就是还剩多少字符要读取，以及read4 api的返回值里较小的一个
- 我们根据这个bytesToRead将read4Buf里的值拷贝到buf里，并且增加bytesRead
- 最后循环结束后，我们返回bytesRead就是我们究竟读取了多少字符。
Java:
Time Complexity – O(n)，Space Complexity – O(1)
/* The read4 API is defined in the parent class Reader4.
int read4(char[] buf); */public class Solution extends Reader4 {
/**
* @param buf Destination buffer
* @param n Maximum number of characters to read
* @return The number of characters read
*/
public int read(char[] buf, int n) {
char[] read4Buf = new char[4];
int bytesRead = 0;
boolean EOF = false;
while (!EOF && bytesRead <= n) {
int read4Bytes = read4(read4Buf);
if (read4Bytes < 4) {
EOF = true;
}
int bytesToRead = Math.min(n - bytesRead, read4Bytes);
for (int i = 0; i < bytesToRead; i++) {
buf[bytesRead++] = read4Buf[i];
}
}
return bytesRead;
}
}
三刷:
前面写得比较麻烦，这次换了新写法。我们首先创建一个read4Buf用来保存每次调用read4 api所返回的字符。建立两个变量，一个read4Count用来记录read4调用实际返回了多少个字符，另外一个totalCharRead表示我们总共已经读取了多少字符。使用一个while循环，在(read4Count ＝ read4(read4Buf)) > 0的时候，每次拷贝这回read4调用读取的字符到输出buf中。注意拷贝时的条件是i < read4Count && totalCharRead < n
Java:
/* The read4 API is defined in the parent class Reader4.
int read4(char[] buf); */public class Solution extends Reader4 {
/**
* @param buf Destination buffer
* @param n Maximum number of characters to read
* @return The number of characters read
*/
public int read(char[] buf, int n) {
char[] read4Buf = new char[4];
int totalCharRead = 0;
int read4Count = 0;
while ((read4Count = read4(read4Buf)) > 0) {
for (int i = 0; i < read4Count && totalCharRead < n; i++) {
buf[totalCharRead++] = read4Buf[i];
}
}
return totalCharRead;
}
}
Reference：
https://leetcode.com/discuss/19573/accepted-clean-java-solution
