[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)

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:

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:

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++:

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] 158. Read N Characters Given Read4 II - Call multiple times

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