[LeetCode] 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.

Thought process:
This question is different from 157. Read N Characters Given Read4, because if the characters read by read4 is more than n, I must start from the unused extra characters in the temp buffer next time. So the idea is to make temp buffer an instance variable. Use a temp pointer and a temp count to keep track of where and how many to read from temp buffer.

Solution:

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/* The read4 API is defined in the parent class Reader4. int read4(char[] buf); */ public class Solution extends Reader4 { private char[] temp = new char[4]; private int tempPointer = 0; private int tempCount = 0; /** * @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 pointer = 0; while (pointer < n) { if (tempPointer == 0) { tempCount = read4(temp); } if (tempCount == 0) { break; } while (pointer < n && tempPointer < tempCount) { buf[pointer] = temp[tempPointer]; pointer++; tempPointer++; } if (tempPointer == tempCount) { tempPointer = 0; } } return pointer; } }

Time complexity: O(min(n, remaining file)) for every call to read.