LeetCode 158. Read N Characters Given Read4 II - Call multiple times
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:
1 /** 2 * The read4 API is defined in the parent class Reader4. 3 * int read4(char[] buf); 4 */ 5 public class Solution extends Reader4 { 6 LinkedList<Character> que = new LinkedList<>(); 7 8 /** 9 * @param buf Destination buffer 10 * @param n Number of characters to read 11 * @return The number of actual characters read 12 */ 13 public int read(char[] buf, int n) { 14 int readSum = 0; 15 // 先用queue中剩余的上次结果加到buf中 16 while(readSum < n && !que.isEmpty()){ 17 buf[readSum++] = que.poll(); 18 } 19 20 // 若是不够再调用read4 API 21 boolean eof = false; 22 char [] temp = new char[4]; 23 while(!eof && readSum < n){ 24 int count = read4(temp); 25 eof = count < 4; 26 int rest = n-readSum; 27 28 int i = 0; 29 while(i < count && i < rest){ 30 buf[readSum++] = temp[i++]; 31 } 32 33 // 把当前read4Buff中没有读的有用char加到queue中 34 if(i == rest){ 35 while(i < count){ 36 que.add(temp[i++]); 37 } 38 } 39 } 40 41 return readSum; 42 } 43 }
