[LeetCode] 158. 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.

157. Read N Characters Given Read4 的拓展，之前只能调用一次，而这里可以调用多次，又多了一些corner case：

第一次调用时，如果read4读出的多余字符我们要先将其暂存起来，这样第二次调用时先读取这些暂存的字符

第二次调用时，如果连暂存字符都没读完，那么这些暂存字符还得留给第三次调用时使用

所以，难点就在于怎么处理这个暂存字符。因为用数组和指针控制对第二种情况比较麻烦，且这些字符满足先进先出，所以我们可以用一个队列暂存这些字符。这样，只要队列不为空，就先读取队列。

Java: Time: O(n), Space: O(1)

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public class Solution extends Reader4 {     Queue<Character> remain = new LinkedList<Character>();           public int read(char[] buf, int n) {         int i = 0;         // 队列不为空时，先读取队列中的暂存字符         while(i < n && !remain.isEmpty()){             buf[i] = remain.poll();             i++;         }         for(; i < n; i += 4){             char[] tmp = new char[4];             int len = read4(tmp);             // 如果读到字符多于我们需要的字符，需要暂存这些多余字符             if(len > n - i){                 System.arraycopy(tmp, 0, buf, i, n - i);                 // 把多余的字符存入队列中                 for(int j = n - i; j < len; j++){                     remain.offer(tmp[j]);                 }             // 如果读到的字符少于我们需要的字符，直接拷贝             } else {                 System.arraycopy(tmp, 0, buf, i, len);             }             // 同样的，如果读不满4个，说明数据已经读完，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);         }         // 如果到这里，说明都是完美读取，直接返回n         return n;     } }

Python:

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class Solution(object):     def __init__(self):         self.__buf4 = [''] * 4         self.__i4 = 0         self.__n4 = 0       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)         """         i = 0         while i < n:             if self.__i4 < self.__n4:  # Any characters in buf4.                 buf[i] = self.__buf4[self.__i4]                 i += 1                 self.__i4 += 1             else:                 self.__n4 = read4(self.__buf4)  # Read more characters.                 if self.__n4:                     self.__i4 = 0                 else:  # Buffer has been empty.                     break           return i   if __name__ == "__main__":     global file_content     sol = Solution()     buf = ['' for _ in xrange(100)]     file_content = "ab"     print(buf[:sol.read(buf, 1)])     print(buf[:sol.read(buf, 2)])

Python:

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from Queue import Queue   class Solution(object):     def __init__(self):         #self.curTotal = 0         self.buffer = Queue()         self.endOfFile = False     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)         """         if n == 0:             return 0                   total = 0         while self.buffer.qsize() < n and not self.endOfFile:             temp = [""] * 4             r = read4(temp)             if r < 4:                 self.endOfFile = True             for i in range(r):                 self.buffer.put(temp[i])                       for i in range(min(self.buffer.qsize(), n)):             buf[i] = self.buffer.get()             total += 1                   return total

C++:

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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) {         if(n == 0)             return 0;                       int total = 0;         while(this->buffer.size() < n && !this->endOfFile) {             char* temp = new char[4];             int r = read4(temp);             if(r < 4)                 this->endOfFile = true;             for(int i = 0; i < r; i++)                 this->buffer.push(temp[i]);         }           int l = min((int)this->buffer.size(), n);         for(int i = 0; i < l; i++) {             buf[i] = this->buffer.front();             this->buffer.pop();             total++;         }         return total;     }       private:     queue<char> buffer;     bool endOfFile = false; };

　　

　　　

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[LeetCode] 157. Read N Characters Given Read4 用Read4来读取N个字符