[LeetCode] 158. Read N Characters Given Read4 II - Call multiple times 用Read4來(lái)讀取N個(gè)字符之二 - 多次調(diào)用

Given a file and assume that you can only read the file using a given method read4, implement a method read to read n characters. Your method read may be called multiple times.

Method read4:

The API read4 reads 4 consecutive characters from the file, then writes those characters into the buffer array buf.

The return value is the number of actual characters read.

Note that read4() has its own file pointer, much like FILE *fp in C.

Definition of read4:

    Parameter:  char[] buf
Returns:    int

Note: buf[] is destination not source, the results from read4 will be copied to buf[]

Below is a high level example of how read4 works:

File file("abcdefghijk"); // File is "abcdefghijk", initially file pointer (fp) points to 'a'
char[] buf = new char[4]; // Create buffer with enough space to store characters
read4(buf); // read4 returns 4. Now buf = "abcd", fp points to 'e'
read4(buf); // read4 returns 4. Now buf = "efgh", fp points to 'i'
read4(buf); // read4 returns 3. Now buf = "ijk", fp points to end of file

Method read:

By using the read4 method, implement the method read that reads n characters from the file and store it in the buffer array buf. Consider that you cannot manipulate the file directly.

The return value is the number of actual characters read.

Definition of read:

    Parameters: char[] buf, int n
Returns: int

Note: buf[] is destination not source, you will need to write the results to buf[]

Example 1:

File file("abc");
Solution sol;
// Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.
sol.read(buf, 1); // After calling your read method, buf should contain "a". We read a total of 1 character from the file, so return 1.
sol.read(buf, 2); // Now buf should contain "bc". We read a total of 2 characters from the file, so return 2.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.

Example 2:

File file("abc");
Solution sol;
sol.read(buf, 4); // After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.

Note:

1. Consider that you cannot manipulate the file directly, the file is only accesible for read4 but not for read.
2. The read function may be called multiple times.
3. Please remember to RESET your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see here for more details.
4. You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.
5. It is guaranteed that in a given test case the same buffer buf is called by read.

這道題是之前那道 Read N Characters Given Read4 的拓展，那道題說(shuō) read 函數(shù)只能調(diào)用一次，而這道題說(shuō) read 函數(shù)可以調(diào)用多次，那么難度就增加了，為了更簡(jiǎn)單直觀的說(shuō)明問(wèn)題，舉個(gè)簡(jiǎn)單的例子吧，比如：

buf = "ab", [read(1),read(2)]，返回 ["a","b"]

那么第一次調(diào)用 read(1) 后，從 buf 中讀出一個(gè)字符，就是第一個(gè)字符a，然后又調(diào)用了一個(gè) read(2)，想取出兩個(gè)字符，但是 buf 中只剩一個(gè)b了，所以就把取出的結(jié)果就是b。再來(lái)看一個(gè)例子：

buf = "a", [read(0),read(1),read(2)]，返回 ["","a",""]

第一次調(diào)用 read(0)，不取任何字符，返回空，第二次調(diào)用 read(1)，取一個(gè)字符，buf 中只有一個(gè)字符，取出為a，然后再調(diào)用 read(2)，想取出兩個(gè)字符，但是 buf 中沒(méi)有字符了，所以取出為空。

但是這道題我不太懂的地方是明明函數(shù)返回的是 int 類型啊，為啥 OJ 的 output 都是 vector<char> 類的，然后我就在網(wǎng)上找了下面兩種能通過(guò)OJ的解法，大概看了看，也是看的個(gè)一知半解，貌似是用兩個(gè)變量 readPos 和 writePos 來(lái)記錄讀取和寫的位置，i從0到n開(kāi)始循環(huán)，如果此時(shí)讀和寫的位置相同，那么調(diào)用 read4 函數(shù)，將結(jié)果賦給 writePos，把 readPos 置零，如果 writePos 為零的話，說(shuō)明 buf 中沒(méi)有東西了，返回當(dāng)前的坐標(biāo)i。然后用內(nèi)置的 buff 變量的 readPos 位置覆蓋輸入字符串 buf 的i位置，如果完成遍歷，返回n，參見(jiàn)代碼如下：

解法一：

// Forward declaration of the read4 API.
int read4(char *buf);

class Solution {
public:
    int read(char *buf, int n) {
        for (int i = 0; i < n; ++i) {
            if (readPos == writePos) {
                writePos = read4(buff);
                readPos = 0;
                if (writePos == 0) return i;
            }
            buf[i] = buff[readPos++];
        }
        return n;
    }
private:
    int readPos = 0, writePos = 0;
    char buff[4];
};

下面這種方法和上面的方法基本相同，稍稍改變了些解法，使得看起來(lái)更加簡(jiǎn)潔一些：

解法二：

// Forward declaration of the read4 API.
int read4(char *buf);

class Solution {
public:
    int read(char *buf, int n) {
        int i = 0;
        while (i < n && (readPos < writePos || (readPos = 0) < (writePos = read4(buff))))
            buf[i++] = buff[readPos++];
        return i;
    }
    char buff[4];
    int readPos = 0, writePos = 0;
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/158

類似題目：

Read N Characters Given Read4

參考資料：

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

https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/discuss/49598/A-simple-Java-code

https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/discuss/49607/The-missing-clarification-you-wish-the-question-provided

到此這篇關(guān)于C++實(shí)現(xiàn)LeetCode(158.用Read4來(lái)讀取N個(gè)字符之二 - 多次調(diào)用)的文章就介紹到這了,更多相關(guān)C++實(shí)現(xiàn)用Read4來(lái)讀取N個(gè)字符之二 - 多次調(diào)用內(nèi)容請(qǐng)搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持腳本之家！

最新評(píng)論