Leetcode

Read N Characters Given read4 II - Call Multiple Times

Approach 1: $O(2(R + W))$

  • Time:O(n)
  • Space:O(1)

C++

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char *buf4);
 */

class Solution {
 public:
  /**
   * @param buf Destination buffer
   * @param n   Number of characters to read
   * @return    The number of actual characters read
   */
  int read(char* buf, int n) {
    int i = 0;  // buf's index

    while (i < n) {
      if (i4 == n4) {      // all characters in buf4 are consumed
        i4 = 0;            // reset buf4's index
        n4 = read4(buf4);  // read 4 (or less) chars from file to buf4
        if (n4 == 0)       // reach the EOF
          return i;
      }
      buf[i++] = buf4[i4++];
    }

    return i;
  }

 private:
  char* buf4 = new char[4];
  int i4 = 0;  // buf4's index
  int n4 = 0;  // buf4's size
};

JAVA

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char[] buf4);
 */

public class Solution extends Reader4 {
  /**
   * @param buf Destination buffer
   * @param n   Number of characters to read
   * @return    The number of actual characters read
   */
  public int read(char[] buf, int n) {
    int i = 0; // buf's index

    while (i < n) {
      if (i4 == n4) {     // all characters in buf4 are consumed
        i4 = 0;           // reset buf4's index
        n4 = read4(buf4); // read 4 (or less) chars from file to buf4
        if (n4 == 0)      // reach the EOF
          return i;
      }
      buf[i++] = buf4[i4++];
    }

    return i;
  }

  private char[] buf4 = new char[4];
  private int i4 = 0; // buf4's index
  private int n4 = 0; // buf4's size
}

Python

# The read4 API is already defined for you.
# def read4(buf4: List[str]) -> int:

class Solution:
  def read(self, buf: List[str], n: int) -> int:
    i = 0  # buf's index

    while i < n:
      if self.i4 == self.n4:  # all characters in buf4 are consumed
        self.i4 = 0  # reset buf4's index
        self.n4 = read4(self.buf4)  # read 4 (or less) chars from file to buf4
        if self.n4 == 0:  # reach the EOF
          return i
      buf[i] = self.buf4[self.i4]
      i += 1
      self.i4 += 1

    return i

  buf4 = [' '] * 4
  i4 = 0  # buf4's index
  n4 = 0  # buf4's size

Approach 2: $O(R + W)$

  • Time:O(n)
  • Space:O(1)

C++

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char *buf4);
 */

class Solution {
 public:
  /**
   * @param buf Destination buffer
   * @param n   Number of characters to read
   * @return    The number of actual characters read
   */
  int read(char* buf, int n) {
    int i = 0;  // buf's index

    // put remaining chars in buf4 to buf
    while (i4 < n4 && i < n)
      buf[i++] = buf4[i4++];

    // while not reaching the tail (< 4 chars)
    while (i + 4 < n) {
      const int k = read4(buf + i);  // directly write to buf
      if (k == 0)                    // reach the EOF
        return i;
      i += k;
    }

    // reach the tail
    while (i < n) {
      if (i4 == n4) {      // all characters in buf4 are consumed
        i4 = 0;            // reset buf4's index
        n4 = read4(buf4);  // read 4 (or less) chars from file to buf4
        if (n4 == 0)       // reach the EOF
          return i;
      }
      buf[i++] = buf4[i4++];
    }

    return i;
  }

 private:
  char* buf4 = new char[4];
  int i4 = 0;  // buf4's index
  int n4 = 0;  // buf4's size
};