LeetCode 158. Read N Characters Given Read4 II

原题链接在这里：https://leetcode.com/problems/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.

题解：

需要多次调用，用queue来保存前一次调用read4没用完的数据.

read时先用queue中的数据添加到buf中，若是不够再call read4. 

在读够n个char后若是read4Buff中还有可用数据，加到queue中. 

Note: declear rest first, but not use i < n - readSum in the while condidtion since readSum is changing.

Time Complexity: read, O(n).

Space: O(1). queue的大小不会超过4.

AC Java:

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char[] buf); 
 */
public class Solution extends Reader4 {
    LinkedList<Character> que = new LinkedList<>();
    
    /**
     * @param buf Destination buffer
     * @param n   Number of characters to read
     * @return    The number of actual characters read
     */
    public int read(char[] buf, int n) {
        int readSum = 0;
        // 先用queue中剩余的上次结果加到buf中
        while(readSum < n && !que.isEmpty()){
            buf[readSum++] = que.poll();
        }
        
        // 若是不够再调用read4 API
        boolean eof = false;
        char [] temp = new char[4];
        while(!eof && readSum < n){
            int count = read4(temp);
            eof = count < 4;
            int rest = n-readSum;
            
            int i = 0;
            while(i < count && i < rest){
                buf[readSum++] = temp[i++];
            }
            
            // 把当前read4Buff中没有读的有用char加到queue中
            if(i == rest){
                while(i < count){
                    que.add(temp[i++]);
                }
            }
        }
        
        return readSum;
    }
}

类似Read N Characters Given Read4.