[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 = 0, slen = 0;
        //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 + 4 - 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 = 0, 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 + 4 - slen);
                cur += n + 4 - 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

posted @ 2016-02-27 15:41  CarlGoodman  阅读(174)  评论(0编辑  收藏  举报