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C++實(shí)現(xiàn)LeetCode(157.用Read4來(lái)讀取N個(gè)字符)

 更新時(shí)間:2021年07月30日 14:37:42   作者:Grandyang  
這篇文章主要介紹了C++實(shí)現(xiàn)LeetCode(157.用Read4來(lái)讀取N個(gè)字符),本篇文章通過(guò)簡(jiǎn)要的案例,講解了該項(xiàng)技術(shù)的了解與使用,以下就是詳細(xì)內(nèi)容,需要的朋友可以參考下

[LeetCode] 157. Read N Characters Given Read4 用Read4來(lái)讀取N個(gè)字符

Given a file and assume that you can only read the file using a given method read4, implement a method to read n characters.

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:

Input: file = "abc", n = 4
Output: 3
Explanation: After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3. Note that "abc" is the file's content, not buf. buf is the destination buffer that you will have to write the results to.

Example 2:

Input: file = "abcde", n = 5
Output: 5
Explanation: After calling your read method, buf should contain "abcde". We read a total of 5 characters from the file, so return 5.

Example 3:

Input: file = "abcdABCD1234", n = 12
Output: 12
Explanation: After calling your read method, buf should contain "abcdABCD1234". We read a total of 12 characters from the file, so return 12.

Example 4:

Input: file = "leetcode", n = 5
Output: 5
Explanation: After calling your read method, buf should contain "leetc". We read a total of 5 characters from the file, so return 5.

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 will only be called once for each test case.
  3. You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.

這道題給了我們一個(gè) Read4 函數(shù),每次可以從一個(gè)文件中最多讀出4個(gè)字符,如果文件中的字符不足4個(gè)字符時(shí),返回準(zhǔn)確的當(dāng)前剩余的字符數(shù)。現(xiàn)在讓實(shí)現(xiàn)一個(gè)最多能讀取n個(gè)字符的函數(shù)。這題有迭代和遞歸的兩種解法,先來(lái)看迭代的方法,思路是每4個(gè)讀一次,然后把讀出的結(jié)果判斷一下,如果為0的話,說(shuō)明此時(shí)的 buf 已經(jīng)被讀完,跳出循環(huán),直接返回 res 和n之中的較小值。否則一直讀入,直到讀完n個(gè)字符,循環(huán)結(jié)束,最后再返回 res 和n之中的較小值,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) {
        int res = 0;
        for (int i = 0; i <= n / 4; ++i) {
            int cur = read4(buf + res);
            if (cur == 0) break;
            res += cur;
        }
        return min(res, n);
    }
};

下面來(lái)看遞歸的解法,這個(gè)也不難,對(duì) buf 調(diào)用 read4 函數(shù),然后判斷返回值t,如果返回值t大于等于n,說(shuō)明此時(shí)n不大于4,直接返回n即可,如果此返回值t小于4,直接返回t即可,如果都不是,則直接返回調(diào)用遞歸函數(shù)加上4,其中遞歸函數(shù)的 buf 應(yīng)往后推4個(gè)字符,此時(shí)n變成n-4即可,參見(jiàn)代碼如下:

解法二:

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

class Solution {
public:
    int read(char *buf, int n) {
        int t = read4(buf);
        if (t >= n) return n;
        if (t < 4) return t;
        return 4 + read(&buf[4], n - 4);
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/157

類似題目:

Read N Characters Given Read4 II - Call multiple times

參考資料:

https://leetcode.com/problems/read-n-characters-given-read4/

https://leetcode.com/problems/read-n-characters-given-read4/discuss/49557/Accepted-clean-java-solution

https://leetcode.com/problems/read-n-characters-given-read4/discuss/49496/Another-accepted-Java-solution

到此這篇關(guān)于C++實(shí)現(xiàn)LeetCode(157.用Read4來(lái)讀取N個(gè)字符)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!

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