Given a file and assume that you can only read the file using a given method read4
, implement a method read
to read n characters. Your method read
may be called multiple times.
Method read4:
The API read4
reads 4 consecutive characters from the file, then writes those characters into the buffer array buf
.
The return value is the number of actual characters read.
Note that read4()
has its own file pointer, much like FILE *fp
in C.
Definition of read4:
Parameter: char[] buf
Returns: int
Note: buf[] is destination not source, the results from read4 will be copied to buf[]
Below is a high level example of how read4
works:
File file("abcdefghijk"); // File is "abcdefghijk", initially file pointer (fp) points to 'a'
char[] buf = new char[4]; // Create buffer with enough space to store characters
read4(buf); // read4 returns 4. Now buf = "abcd", fp points to 'e'
read4(buf); // read4 returns 4. Now buf = "efgh", fp points to 'i'
read4(buf); // read4 returns 3. Now buf = "ijk", fp points to end of file
Method read:
By using the read4
method, implement the method read
that reads n characters from the file and store it in the buffer array buf
. Consider that you cannot manipulate the file directly.
The return value is the number of actual characters read.
Definition of read:
Parameters: char[] buf, int n
Returns: int
Note: buf[] is destination not source, you will need to write the results to buf[]
Example 1:
File file("abc");
Solution sol;
// Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.
sol.read(buf, 1); // After calling your read method, buf should contain "a". We read a total of 1 character from the file, so return 1.
sol.read(buf, 2); // Now buf should contain "bc". We read a total of 2 characters from the file, so return 2.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
Example 2:
File file("abc");
Solution sol;
sol.read(buf, 4); // After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
Note:
- Consider that you cannot manipulate the file directly, the file is only accesible for
read4
but not forread
. - The
read
function may be called multiple times. - Please remember to RESET your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see here for more details.
- You may assume the destination buffer array,
buf
, is guaranteed to have enough space for storing n characters. - It is guaranteed that in a given test case the same buffer
buf
is called byread
.
这道题是之前那道 Read N Characters Given Read4 的拓展,那道题说 read 函数只能调用一次,而这道题说 read 函数可以调用多次,那么难度就增加了,为了更简单直观的说明问题,举个简单的例子吧,比如:
buf = “ab”, [read(1),read(2)],返回 [“a”,”b”]
那么第一次调用 read(1) 后,从 buf 中读出一个字符,就是第一个字符a,然后又调用了一个 read(2),想取出两个字符,但是 buf 中只剩一个b了,所以就把取出的结果就是b。再来看一个例子:
buf = “a”, [read(0),read(1),read(2)],返回 [“”,”a”,””]
第一次调用 read(0),不取任何字符,返回空,第二次调用 read(1),取一个字符,buf 中只有一个字符,取出为a,然后再调用 read(2),想取出两个字符,但是 buf 中没有字符了,所以取出为空。
但是这道题我不太懂的地方是明明函数返回的是 int 类型啊,为啥 OJ 的 output 都是 vector
解法一:
// Forward declaration of the read4 API.
int read4(char *buf);
class Solution {
public:
int read(char *buf, int n) {
for (int i = 0; i < n; ++i) {
if (readPos == writePos) {
writePos = read4(buff);
readPos = 0;
if (writePos == 0) return i;
}
buf[i] = buff[readPos++];
}
return n;
}
private:
int readPos = 0, writePos = 0;
char buff[4];
};
下面这种方法和上面的方法基本相同,稍稍改变了些解法,使得看起来更加简洁一些:
解法二:
// Forward declaration of the read4 API.
int read4(char *buf);
class Solution {
public:
int read(char *buf, int n) {
int i = 0;
while (i < n && (readPos < writePos || (readPos = 0) < (writePos = read4(buff))))
buf[i++] = buff[readPos++];
return i;
}
char buff[4];
int readPos = 0, writePos = 0;
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/158
类似题目:
参考资料:
https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/
LeetCode All in One 题目讲解汇总(持续更新中…)
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