LeetCode Read N Characters Given Read4

原题链接在这里：https://leetcode.com/problems/read-n-characters-given-read4/

题目：

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.

题解：

Given API read4, 一次最多可以read 4个char, 并把这些char保留在temp Buff中.

Ask to design another API, 能一次最多读n个char. 每次call read4, 若是返回小于4, 说明已经到了end of file, 下一次跳出while loop.

readCount计数当前共读了多少char, n-readCount就是还需要读多少个char.

Time Complexity: O(n). Space: O(1).

AC Java:

 /* 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) {
         boolean eof = false;
         char [] temp = new char[4];
         int readCount = 0;
         while(!eof && readCount<n){
             int count = read4(temp);    //read4 API读的byte都存在temp Buff中
             eof = count < 4;

             count = Math.min(count, n-readCount);   //如果n减掉已经读的char的数目比count小，就只读n-readCount个char
             for(int i = 0; i<count; i++){
                 buf[readCount++] = temp[i];
             }
         }
         return readCount;
     }
 }

跟上Read N Characters Given Read4 II - Call multiple times.