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