Given a file and assume that you can only read the file using a given method read4, implement a method to read n characters.
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:
Input: file = "abc", n = 4 Output: 3 Explanation: After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3. Note that "abc" is the file's content, not buf. buf is the destination buffer that you will have to write the results to.
Example 2:
Input: file = "abcde", n = 5 Output: 5 Explanation: After calling your read method, buf should contain "abcde". We read a total of 5 characters from the file, so return 5.
Example 3:
Input: file = "abcdABCD1234", n = 12 Output: 12 Explanation: After calling your read method, buf should contain "abcdABCD1234". We read a total of 12 characters from the file, so return 12.
Example 4:
Input: file = "leetcode", n = 5 Output: 5 Explanation: After calling your read method, buf should contain "leetc". We read a total of 5 characters from the file, so return 5.
Note:
- Consider that you cannot manipulate the file directly, the file is only accesible for
read4but not forread. - The
readfunction will only be called once for each test case. - You may assume the destination buffer array,
buf, is guaranteed to have enough space for storing n characters.
這道題給了我們一個 Read4 函數，每次可以從一個文件中最多讀出4個字符，如果文件中的字符不足4個字符時，返回准確的當前剩余的字符數。現在讓實現一個最多能讀取n個字符的函數。這題有迭代和遞歸的兩種解法，先來看迭代的方法，思路是每4個讀一次，然后把讀出的結果判斷一下，如果為0的話，說明此時的 buf 已經被讀完，跳出循環，直接返回 res 和n之中的較小值。否則一直讀入，直到讀完n個字符，循環結束，最后再返回 res 和n之中的較小值，參見代碼如下：
解法一：
// Forward declaration of the read4 API. int read4(char *buf); class Solution { public: int read(char *buf, int n) { int res = 0; for (int i = 0; i <= n / 4; ++i) { int cur = read4(buf + res); if (cur == 0) break; res += cur; } return min(res, n); } };
下面來看遞歸的解法，這個也不難，對 buf 調用 read4 函數，然后判斷返回值t，如果返回值t大於等於n，說明此時n不大於4，直接返回n即可，如果此返回值t小於4，直接返回t即可，如果都不是，則直接返回調用遞歸函數加上4，其中遞歸函數的 buf 應往后推4個字符，此時n變成n-4即可，參見代碼如下：
解法二：
// Forward declaration of the read4 API. int read4(char *buf); class Solution { public: int read(char *buf, int n) { int t = read4(buf); if (t >= n) return n; if (t < 4) return t; return 4 + read(&buf[4], n - 4); } };
Github 同步地址：
https://github.com/grandyang/leetcode/issues/157
類似題目：
Read N Characters Given Read4 II - Call multiple times
參考資料：
https://leetcode.com/problems/read-n-characters-given-read4/