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Leetcode: 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.Show TagsHave you met this question in a real interview?
这道题跟I不一样在于，read函数可能多次调用，比如read(buf,23)之后又read(buf, 25), 第一次调用时的buffer还没用完，还剩一个char在buffer里，第二次拿出来接着用，这样才能保证接着上次读的地方继续往下读。
1. 所以应该设置这4个char的buffer为instance variable（实例变量），这样每次调用read函数后buffer还剩的元素可以被保存下来，供给下一次读取
2. 那么下一次调用read函数时候，怎么知道上一次buffer里面还剩不剩未读元素呢？我们有oneRead(一次读到buffer里的char数)，actRead(实际被读取的char数)，oneRead-actRead就是还剩在buffer里的char数。通常oneRead == actRead, 只有当n-haveRead < oneRead时，才不等，这就是上一次调用read结束的时候。所以只要调用read函数发现oneRead != 0 时，就说明上一次调用read还剩了元素在buffer里，先读完这个，再调用read4读新的。oneRead也需要是instance varaible
3. 还需要设置一个offset: Use to keep track of the offset index where the data begins in the nextread call. The buffer could be read partially (due to constraints of reading upto n bytes) and therefore leaving some data behind.
|<--buffer-->|
// |_________________________|
// |<---offset---> |<----oneRead--->
上图所示为一次read最后的情况，offset部分其实就是actRead的部分，oneRead = oneRead - actRead, 就剩下了右边一部分在buffer里没有读，下一次read函数调用，发现oneRead>0, 说明上一次read还剩了一部分没有读
1 /* The read4 API is defined in the parent class Reader4. 2 int read4(char[] buf); */ 3 4 public class Solution extends Reader4 { 5 /** 6 * @param buf Destination buffer 7 * @param n Maximum number of characters to read 8 * @return The number of characters read 9 */10 private char buffer = new char[4];11 private int oneRead = 0;12 private int offset = 0;13 14 public int read(char[] buf, int n) {15 boolean lessthan4 = false;16 int haveRead = 0;17 while (!lessthan4 && haveRead < n) {18 if (oneRead == 0) {19 oneRead = read4(buffer);20 lessthan4 = oneRead < 4;21 }22 int actRead = Math.min(n-haveRead, oneRead);23 for (int i=0; i<actRead; i++) {24 buf[haveRead+i] = buffer[offset+i];25 }26 oneRead -= actRead;27 offset = (offset + actRead) % 4;28 haveRead += actRead;29 }30 return haveRead;31 }32 }
Leetcode: Read N Characters Given Read4 II - Call multiple times