157. Read N Characters Given Read4
Total Accepted: 17410 Total Submissions: 58870 Difficulty: EasyThe 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.
思路：
一开始没看懂题。看了好几个人的blog才知道是怎么回事。首先输入数组的长度我们是不知道的。
read4这个API中的参数是读了的char，也就是每次read4之后返回的char[4]是从给定文件中读的。
然后为啥是两参数？char[] buf和上面的一个道理，是输出的数组，而不是源文件；给int n是因为这个文件的长度可能比我们给的大，这样我们只需要读入N个字符到 char[] buf中；也有可能短，那么我们只能读完所有的文件并且返回文件长度。
/* 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) {
int count = 0;
char[] mybuf = new char[4];
while(count<n){
int num = read4(mybuf);
if(num == 0) break;
int index = 0;
while(index < num && count < n){
buf[count++] = mybuf[index++];
}
}
return count;
}
}
// case 1 有数据，n个了; while循环管
// case 2 数据不够n个，read4返回了0。 有一行代码管
158. Read N Characters Given Read4 II - Call multiple times
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 may be called multiple times.
思路：
又是一开始没看懂题。变化是这样：比如先call了n=3，然后call n=5，那么第一次就读入了4个char，第二次call应该把上一次的最后一个char拿来。也就是说要有个cache取缓存已读取的字符，然后从这个cache里面取。
每次读4个字符，放入 cache，然后
in = Math.min(n - total, buff.size()) 这个是精髓，这样就知道还需要多少个字符了。如果是in=0了，那么说明已经够了或者没字符了，这样就不用再读取了。否则无论哪个大，接下来都需要往buf中放入in个字符（可能是剩的少，也有可能要求的少）。
/* 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
*/
LinkedList<Character> buff = new LinkedList<Character>();
public int read(char[] buf, int n) {
int total = 0;
while(true){
char[] temp = new char[4];
int in = read4(temp);
for(int i = 0; i < in; i++) buff.add(temp[i]);
// 判断还需要写入多少个到结果， 比如读了4个，但是只要3个；或者要4个，只有3个了。
in = Math.min(n - total, buff.size());
for(int i = 0; i < in; i++) buf[total++] = buff.poll();
if(in == 0) break;
}
return total;
}
}