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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 will only be called once for each test case.
刚看题的时候，看了好久才明白到底是怎么一回事
* @param buf Destination buffer，我后来理解这个应该是个足够大的buffer，需要将读到的file放到这个buffer里面去
* @param n Maximum number of characters to read，想要读n个char，但是有可能读不到那么多，因为有可能n比文件的size大
* @return The number of characters read，实际读到的char数，小于等于n
由于不知道文件的大小，所以只能一次一次用4个char的buffer用read4去读，存在以下两种结束方式（假设文件大小为size）：
1. n >> size, 比如n = 50, size = 23, 最后一次读的大小为<4, 下一次就不读了，然后把这一次读的char数oneRead赋给目标buf
至于n = 50, size = 20, 最后一次读到0个，不会进行赋值，下一次也不读了
2. n << size，比如n = 23, size = 50, read4每次会读满4个，但是我们只取3个，取零头的方法是int actRead = Math.min(n-haveRead, oneRead); 下一次OneRead+haveRead>n, 也就停止了
想清楚了这些，就可以写了
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 public int read(char[] buf, int n) { 11 char[] buffer = new char[4]; 12 int haveRead = 0; 13 boolean lessthan4 = false; 14 15 while (!lessthan4 && haveRead < n) { 16 int oneRead = read4(buffer); 17 if (oneRead < 4) lessthan4 = true; 18 int actRead = Math.min(n-haveRead, oneRead); 19 while (int i=1; i<=actRead; i++) { 20 buf[haveRead+i-1] = buffer[i-1]; 21 } 22 haveRead += actRead; 23 } 24 return haveRead; 25 } 26 }
Leetcode: Read N Characters Given Read4
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原文地址：http://www.cnblogs.com/EdwardLiu/p/4233533.html
