[LeetCode] 157. Read N Characters Given Read4 用Read4来读取N个字符
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 will only be called once for each test case.
给一个Read4函数,每次可以从一个文件中最多读出4个字符,如果文件中的字符不足4个,返回当前剩余的字符。实现一个读取n个字符的函数。
用一个临时数组,存放每次read4读到字符,再用一个指针标记buf数组目前存储到的位置,然后将这个临时数组的内容存到buf相应的位置就行了。
需要注意两个corner case:
如果本次读到多个字符,但是我们只需要其中一部分就能完成读取任务时,我们要拷贝的长度是本次读到的个数和剩余所需个数中较小的
如果read4没有读满4个,说明数据已经读完,这时候对于读到的数据长度,因为也可能存在我们只需要其中一部分的情况,所以要返回总所需长度和目前已经读到的长度的较小的
Java: Time: O(n), Space: O(1)
public class Solution extends Reader4 { public int read(char[] buf, int n) { for(int i = 0; i < n; i += 4){ char[] tmp = new char[4]; // 将数据读入临时数组 int len = read4(tmp); // 将临时数组拷贝至buf数组,这里拷贝的长度是本次读到的个数和剩余所需个数中较小的 System.arraycopy(tmp, 0, buf, i, Math.min(len, n - i)); // 如果读不满4个,说明已经读完了,返回总所需长度和目前已经读到的长度的较小的 if(len < 4) return Math.min(i + len, n); } // 如果循环内没有返回,说明读取的字符是4的倍数 return n; } }
Python:
# The read4 API is already defined for you. # @param buf, a list of characters # @return an integer def read4(buf): global file_content i = 0 while i < len(file_content) and i < 4: buf[i] = file_content[i] i += 1 if len(file_content) > 4: file_content = file_content[4:] else: file_content = "" return i class Solution(object): def read(self, buf, n): """ :type buf: Destination buffer (List[str]) :type n: Maximum number of characters to read (int) :rtype: The number of characters read (int) """ read_bytes = 0 buffer = [''] * 4 for i in xrange(n / 4 + 1): size = read4(buffer) if size: size = min(size, n-read_bytes) buf[read_bytes:read_bytes+size] = buffer[:size] read_bytes += size else: break return read_bytes if __name__ == "__main__": global file_content buf = ['' for _ in xrange(100)] file_content = "a" print(buf[:Solution().read(buf, 9)]) file_content = "abcdefghijklmnop" print(buf[:Solution().read(buf, 9)])
C++:
// Time: O(n) // Space: O(1) int read4(char *buf); class Solution { public: /** * @param buf Destination buffer * @param n Maximum number of characters to read * @return The number of characters read */ int read(char *buf, int n) { int read_bytes = 0; char buffer[4]; for (int i = 0; i <= n / 4; ++i) { if (int size = read4(buffer)) { size = min(size, n - read_bytes); for (int j = 0; j < size; ++j) { buf[read_bytes++] = buffer[j]; } } else { break; } } return read_bytes; } };
相关题目:
[LeetCode] 158. Read N Characters Given Read4 II - Call multiple times