157 Read N Characters Given Read4

157. Read N Characters Given Read4​
1. Question
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 theread4API, implement the functionint read(char *buf, int n)that readsncharacters from the file.
Note:
Thereadfunction will only be called once for each test case.
2. Implementation
思路: 这题要我们利用一个已有的函数read4()实现一个自定义的函数read(), 和read4()不同的是，read()里有一个参数n，定义我们读多少byte的数据，而read4()只会一次读4个byte。这里主要的考点是edge cases. 如果read4返回的值是小于4，说明是end of file。同时为了保证我们读取不超过 n个byte的数据，每次将数据copy到buffer时，我们都要取n - index和size之间较小的那个数。
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/* The read4 API is defined in the parent class Reader4.
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      int read4(char[] buf); */
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​
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public class Solution extends Reader4 {
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    /**
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     * @param buf Destination buffer
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     * @param n   Maximum number of characters to read
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     * @return    The number of characters read
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     */
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    public int read(char[] buf, int n) {
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        boolean eof = false;
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        char[] buffer = new char[4];
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        int index = 0;
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        while (!eof && index < n) {
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            int size = read4(buffer);
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            if (size < 4) {
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                eof = true;
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            }
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            int bytes = Math.min(n - index, size);  
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            for (int i = 0; i < bytes; i++) {
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                buf[index++] = buffer[i];
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            }
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        }
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        return index;
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    }
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}
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3. Time & Space Complexity
时间复杂度O(n), 空间复杂度O(1)

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158 Read N Characters Given Read4 II - Call multiple times

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Contents

157. Read N Characters Given Read4

1. Question

2. Implementation

3. Time & Space Complexity

157. Read N Characters Given Read4​

1. Question

2. Implementation

3. Time & Space Complexity

157. Read N Characters Given Read4