[Locked] Read N Characters Given Read4 & Read N Characters Given Read4 II - Call multiple times

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 readsn characters from the file.

Note:
The read function will only be called once for each test case.

分析：

　　read函数被调用一次，那么就直接用代码解释这题即可吧。

代码：

int read4(char *buf);
class Solution {
public:
    int read(char *buf, int n) {
        char *cur = buf;
        int clen = , slen = ;
        //当还有字符可以读出来时
        while((clen = read4(cur))) {
            slen += clen;
            //当字符数目超出n时，只留下n个
            if(slen >= n) {
                cur += n +  - slen;
                break;
            }
            cur += clen;
        }
        *cur = '\0';
        //当字符数目小于n时，文件就读完了，则返回文件总长；若字符数目大于等于n时，返回n
        return slen < n ? slen : n;
    }
};

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.

分析：

　　相比于I，这题要被调用多次，而大量的文件读取操作时间代价很大，为了解决这个问题，可以用一个缓存存储已经读过的字符，我用一个私有的string作为缓存。

代码：

int read4(char *buf);
class Solution {
private:
    string str;
public:
    Solution() {
        str = "";
    }
    int read(char *buf, int n) {
        //利用缓存除去不必要的操作
        if(n <= str.length()) {
            //将string传给字符串数组
            strncpy(buf, str.c_str(), n);
            return n;
        }
        strcpy(buf, str.c_str());
        char *cur = buf + str.length();
        int clen = , slen = int(str.length());
        //当缓存不够用时，继续在文件里读取字符
        while((clen = read4(cur))) {
            slen += clen;
            //当字符数目超出n时，只留下n个
            if(slen >= n) {
                str.append(cur, n +  - slen);
                cur += n +  - slen;
                break;
            }
            //字符串数组传给string
            str.append(cur, clen);
            cur += clen;
        }
        *cur = '\0';
        //当字符数目小于n时，文件就读完了，则返回文件总长；若字符数目大于等于n时，返回n
        return slen < n ? slen : n;
    }
};

注：本题代码没有验证