[LeetCode新题] Read N Characters Given Read4

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.

思路：

由于每次只能读进4个字符，而n未必是4的倍数。所以一直读到len+4>n时停止。剩下m = n-len < 4 个字符。建立一个新的缓存remain读入4个字符，然后将其中的m个写入buff尾部。要注意在整个读入过程中的每个阶段判断是否已经读完文件。

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// 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 len = 0; int m = INT_MAX; while(len+4<=n) { m = read4(buf+len); len += m; if(m<4) break; } if(len==n || m<4) return len; char *remain = new char[5]; m = min(read4(remain),n-len); for(int i=0; i<m; i++) buf[len++] = remain[i]; delete remain; return len; } };