• [Locked] Read N Characters Given Read4 & Read N Characters Given Read4 II


    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 = '';
            //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 = '';
            //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

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  • 原文地址:https://www.cnblogs.com/littletail/p/5222451.html
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