Read N Characters Given Read4

题目地址：
https://leetcode.com/problems/read-n-characters-given-read4/#/description

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 will only be called once for each test case.

解题思路：
这道题主要是要理解read4这个api的用法。它是将file里面的4个char读出来存到buf里面并且返回读到的char的个数。这里有一个follow up，就是如果有多次调用read function，那么就需要保存之前consumed的char。Think that you have 4 chars "a, b, c, d" in the file, and you want to call your function twice like this:

1. read(buf, 1); // should return 'a'
2. read(buf, 3); // should return 'b, c, d'

All the 4 chars will be consumed in the first call. So the tricky part of this question is how can you preserve the remaining 'b, c, d' to the second call.

代码：

public int read(char[] buf, int n) {
    int p = 0;
    while(p < n){
        char[] temp = new  char[4];
        int k = read4(temp);
        for(int i = 0; i < k && p < n; i++){
            buf[p++] = temp[i];
        }
        if(k < 4){
            return p;
        }
    }
    return p;
}

follow up：

    private int buffPtr = 0;
    private int k = 0;
    char[] temp = new char[4];
    public int read(char[] buf, int n) {
        int p = 0;
        while(p < n){
            if(buffPtr == 0){
                k = read4(temp);
            }
            if(k == 0){
               break;
            }
            while(p < n && buffPtr < k){
               buf[p++] = temp[buffPtr++];
            }
            if(buffPtr >= k){
               buffPtr = 0;
            }
        }
        return p;
    }