用 Read4 读取 N 个字符 II - 多次调用
难度:
标签:
题目描述
代码结果
运行时间: 25 ms, 内存: 16.2 MB
/*
* Problem: Given an API int read4(char[] buf), read4 reads 4 characters at a time from a file.
* The return value is the actual number of characters read. Implement a method int read(char[] buf, int n)
* that uses this API to read n characters.
*
* Approach:
* 1. Use a buffer to store the temporary characters read by read4.
* 2. Use Java Streams to manage and process the reading process efficiently.
* 3. Loop until we have read the desired number of characters or there are no more characters to read.
* 4. Each time, call read4 and copy the characters to the final buffer.
* 5. Adjust the count to ensure we do not read more than required.
*/
import java.util.stream.IntStream;
public class Solution extends Reader4 {
private char[] buffer = new char[4];
private int bufferPtr = 0;
private int bufferCount = 0;
public int read(char[] buf, int n) {
int totalRead = 0;
while (totalRead < n) {
if (bufferPtr == bufferCount) {
bufferCount = read4(buffer);
bufferPtr = 0;
if (bufferCount == 0) break; // End of file reached
}
while (totalRead < n && bufferPtr < bufferCount) {
buf[totalRead++] = buffer[bufferPtr++];
}
}
return totalRead;
}
}
解释
方法:
该题解使用一个缓冲区 self.buf 来存储读取到的字符。当调用 read 方法时,会不断调用 read4 方法,将读取到的字符放入缓冲区,直到读取到足够的字符或文件末尾。然后从缓冲区中取出需要的字符返回给调用者。通过维护缓冲区的起始位置 self.start 和结束位置 self.end,可以实现多次调用 read 方法,每次都能从上一次读取的位置继续读取。
时间复杂度:
O(n)
空间复杂度:
O(n)
代码细节讲解
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在`read`方法中,为什么要使用一个中间数组`buf4`来存储`read4`的结果,而不是直接将结果追加到主缓冲区`self.buf`?
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当`read4`返回的字符少于4个时(即文件末尾),您是如何处理这种情况以防止对缓冲区`self.buf`的越界访问的?
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在`read`方法中,为何需要使用`self.start`和`self.end`来管理缓冲区,这种设计有什么特别的优点吗?
▷🦆
您在代码中使用了`self.end += 4 if n >= 4 else n`这一行进行缓冲区结束位置的更新,这里的逻辑是否能正确处理所有情况,例如当`read4`读取的字符数少于4个时?
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