[Leetcode] 158. Read N Characters Given Read4 II

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.

Note:
The read function may be called multiple times.

思路：

建议读者将这道题目和Leetcode 157结合起来思考。它和Leetcode 157的不同之处就在于read(char *buf, int n)方法可能被调用多次，这样就产生了一个问题：上一次调用时读到buffer中的内容可能没有被取尽，所以下次调用read方法时，需要优先从上次没有读完的buffer中读取数据。所以，我们需要在类中记录三个类变量：长度为4的缓冲区(buffer)，有效长度(last_length)以及起始位置(last_pos)。每次在调用read(char *buf, int n)的时候，我们首先从buffer中读取剩余数据；然后再调用read4这一API从文件中读取数据（这一部分的思路就和Leetcode 157完全一样了）。

代码：

// Forward declaration of the read4 API.int read4(char *buf);class Solution {public:    /**     * @param buf Destination buffer     * @param n   Maximum number of characters to read     * @return    The number of characters read     */    int read(char *buf, int n) {        int ret = 0;          while(last_pos < last_length && n-- > 0) {       // read the left charachters from last time            buf[ret++] = buffer[last_pos++];          }        while(n > 0) {              last_length = read4(buffer);            if(last_length == 0) {                return ret;            }            last_pos = 0;              while(last_pos < last_length && n-- > 0) {   // read the current chacters until read over or the read length reaches n                buf[ret++] = buffer[last_pos++];              }        }          return ret;    }private:    int last_pos = 0, last_length = 0;      char buffer[4];};