LeetCode #158 - Read N Characters Given Read4 II - Call multiple times

最新推荐文章于 2021-08-05 23:01:16 发布

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本文介绍了一种利用read4 API实现read函数的方法，该函数可以从文件中读取指定数量的字符。通过维护一个内部缓冲区和两个状态变量，可以在多次调用中连续读取文件内容，直到达到指定的字符数或文件结束。

题目描述：

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.

Example 1:

Given buf = "abc"

read("abc", 1) // returns "a"

read("abc", 2); // returns "bc"

read("abc", 1); // returns ""

Example 2:

Given buf = "abc"

read("abc", 4) // returns "abc"

read("abc", 1); // returns ""

// 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) {
        for(int i=0;i<n;i++)
        {
            //当write_pos等于read_num时，说明buffer中的字符都已经写入了buf中，需要再从file中读取
            if(write_pos==read_num) 
            {
                read_num=read4(buffer);
                write_pos=0; //write_pos从0开始
                if(read_num==0) return i; //说明file已经读取完，直接返回i
            }
            buf[i]=buffer[write_pos];
            write_pos++;
        }
        return n;
    }

private:
    int write_pos=0; //表示当前需要bffer的哪一位写入buf中
    int read_num=0;//表示当前一共从file中读取了多少字符到buffer中
    char buffer[4];
};