157. Read N Characters Given Read4 (Easy)

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.

Solution 1: iterative

这道题给了我们一个Read4函数，每次可以从一个文件中最多读出4个字符，如果文件中的字符不足4个字符时，返回准确的当前剩余的字符数。现在让我们实现一个最多能读取n个字符的函数。这题有迭代和递归的两种解法，我们先来看迭代的方法，思路是我们每4个读一次，然后把读出的结果判断一下，如果为0的话，说明此时的buf已经被读完，跳出循环，直接返回res和n之中的较小值。否则一直读入，直到读完n个字符，循环结束，最后再返回res和n之中的较小值，参见代码如下：

version 1: 3ms

// 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 res = 0;
        for (int i = 0; i <= n/4; ++i) {
            int cur = read4(buf+res);
            if (cur == 0) break;
            res += cur;
        }
        return min(res, n);
    }
};

version 2: 3ms

// 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 res = 0;
        while (n > 0) {
            int tmp = min(read4(buf),n);
            res += tmp;
            if (tmp < 4) {
                return res;
                break;
            }
            buf += 4;
            n -= 4;
        }
        return res;
    }
};