[leetcode] 158. Read N Characters Given Read4 II - Call multiple times 解题报告

题目链接：https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/

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：题意是说可能调用很多次，也就是说上次可能最后读入了4个，但是只取了2个，还剩了两个。所以要做一个标记，把上次读入了但是没有取走的字符串保留到这次。

我们可以做两个标记，一个上次剩余的长度，还有在数组中的位置，然后这个读入的时候先取上次读入的。

代码如下：

// Forward declaration of the read4 API.
int read4(char *buf);

class Solution {
public:
    Solution():len(0), pos(0){}
    /**
     * @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 tem = 0, offset = 0;
        while(n > 0 && len > 0)
        {
            buf[tem++] = str[pos++];
            len--, n--, offset++;
        }
        while(n >0 && (tem=read4(str)) > 0)
        {
            len = tem>n?(tem-n):0;
            pos = (tem>n)?n:0;
            tem = min(tem, n);
            memcpy(buf+offset, str, tem);
            n -= tem;
            offset += tem;
        }
        return offset;
    }
private:
    int len, pos;
    char str[4];
};