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

题目：

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.

链接：  http://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/

题解：

又是比较难理解题意的一题...call multiple times，举个例子呀， 没例子咋理解。尝试了好几次才弄明白意思。给定read4，跟上一题一样，求可以call multiple times的read()。假如文件字符串是"abc"，我们调用read(1)，应该返回"a"，再调用read(2)，应该返回bc"。这里要注意的是，之前我们再第一次调用的时候，read4就已经读取了"abc"，所以这道题其实就是要在上一题的基础上处理这种情况。解决方法不难，我们可以用一个queue来存储多读取的部分，然后在下次调用read的时候，根据情况判断，先读queue里面上次读剩下的数据，再进行下面的读取。也可以把read4的buffer放在global，然后用一个int型的offset来记录上次读了read4的多少。由于这个global buffer不会超过4，所以space complexity算是O(1)的。 LeetCode有些题真的很难读懂题意。

Time Complexity - O(n)， Space Complexity - O(1)。

/* The read4 API is defined in the parent class Reader4.
      int read4(char[] buf); */

public class Solution extends Reader4 {
    /**
     * @param buf Destination buffer
     * @param n   Maximum number of characters to read
     * @return    The number of characters read
     */
     
    private Queue<Character> q;
    private boolean EOF;
    
    public Solution() {
        this.q = new LinkedList<>();
        this.EOF = false;
    }
    
    public int read(char[] buf, int n) {
        char[] read4Buffer = new char[4];
        int bytesRead = 0;
        
        while (!q.isEmpty() && bytesRead < n)             //try read queue buffer first
            buf[bytesRead++] = q.poll();
        
        while(!this.EOF && bytesRead < n) {
            int read4Bytes = read4(read4Buffer);
            if(read4Bytes < 4)
                this.EOF = true;
            int bytes = Math.min(n - bytesRead, read4Bytes);
            System.arraycopy(read4Buffer, 0, buf, bytesRead, bytes);
            bytesRead += bytes;
            
            if(bytes < 4) {                                 //push read4 reminder to q for next read
                for(int i = bytes; i < read4Bytes; i++)
                    this.q.offer(read4Buffer[i]);
            }
        }
        
        return bytesRead;
    }
}

 

二刷:

这回题意理解得还算比较顺利。就是给一个read4()的api，每次最多读取4个char，要求实现read，读取n个char。这里n可以小于read4()返回的数字，也可以大于。

问题的关键是要储存多读的字符，我们使用一个queue就可以简单解决。每次调用read()的时候，先检查queue是否为空，假如不为空则先从queue中读取。接下来，假如仍然没有读到n个字符，我们就跟上一题一样，调用read4()来不断读取。 要注意多读的字符，我们要保存到queue中。最后返回一共读取的字符数就可以了。

也可以理解为把缓存中的东西持久化。

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

/* The read4 API is defined in the parent class Reader4.
      int read4(char[] buf); */

public class Solution extends Reader4 {
    /**
     * @param buf Destination buffer
     * @param n   Maximum number of characters to read
     * @return    The number of characters read
     */
    Queue<Character> remainingChars = new LinkedList<>(); 
     
    public int read(char[] buf, int n) {
        char[] read4Buf = new char[4];
        int read4Count = 0;
        int totalCharsRead = 0;
        while (remainingChars.size() > 0 && totalCharsRead < n) {
            buf[totalCharsRead++] = remainingChars.poll();
        }
        //if (totalCharsRead == n) return n;
        while ((read4Count = read4(read4Buf)) > 0) {
            int i = 0;
            while (i < read4Count && totalCharsRead < n) {
                buf[totalCharsRead++] = read4Buf[i++];
            }
            while (i < read4Count) {
                remainingChars.offer(read4Buf[i++]);
            }
        }
        return totalCharsRead;
    }
}

 

 

Reference：

https://leetcode.com/discuss/19581/clean-accepted-java-solution

https://leetcode.com/discuss/21219/a-simple-java-code

https://leetcode.com/discuss/21393/finally-get-question-understood-and-ac-by-c

https://leetcode.com/discuss/25200/my-python-40ms-solution

http://www.cnblogs.com/EdwardLiu/p/4240616.html