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.
Analysis:
Since the function can be called multiple times, we need to record the left over content in the buffer of read4, and put them in some place. For the next call, we need to read content from the carry over buffer first.
Solution:
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 11 public char[] carry; 12 public int index; 13 14 public int read(char[] buf, int n) { 15 int left = n; 16 17 //Put carry over into buf, and destroy the carry over array! 18 if (carry!=null){ 19 while (left>0 && index<carry.length){ 20 buf[n-left]=carry[index]; 21 left--; 22 index++; 23 } 24 if (index>=carry.length){ 25 carry=null; 26 index=-1; 27 } 28 } 29 30 char[] tempBuf = new char[4]; 31 while (left>0){ 32 int num = read4(tempBuf); 33 //if the read number is larger then what we need, then we just put the left number of chars into buf. 34 //And put the rest chars into carry array. 35 if (num>left){ 36 carry = new char[num-left]; 37 for (int i=left;i<num;i++) 38 carry[i-left] = tempBuf[i]; 39 index = 0; 40 } 41 42 int end = Math.min(num,left); 43 for (int i=0;i<end;i++){ 44 buf[n-left] = tempBuf[i]; 45 left--; 46 } 47 48 //If reach EOF. 49 if (left>0 && num<4) break; 50 } 51 52 return n-left; 53 54 } 55 }
Leetcode-Read N Characters Given Read4 II
原文：http://www.cnblogs.com/lishiblog/p/4160863.html