c 实现leetcode(158.用read4来读取n个字符之二 -尊龙凯时

[leetcode] 158. read n characters given read4 ii - call multiple times 用read4来读取n个字符之二 - 多次调用

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.

这道题是之前那道 read n characters given read4 的拓展，那道题说 read 函数只能调用一次，而这道题说 read 函数可以调用多次，那么难度就增加了，为了更简单直观的说明问题，举个简单的例子吧，比如：

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

那么第一次调用 read(1) 后，从 buf 中读出一个字符，就是第一个字符a，然后又调用了一个 read(2)，想取出两个字符，但是 buf 中只剩一个b了，所以就把取出的结果就是b。再来看一个例子：

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

第一次调用 read(0)，不取任何字符，返回空，第二次调用 read(1)，取一个字符，buf 中只有一个字符，取出为a，然后再调用 read(2)，想取出两个字符，但是 buf 中没有字符了，所以取出为空。

但是这道题我不太懂的地方是明明函数返回的是 int 类型啊，为啥 oj 的 output 都是 vector 类的，然后我就在网上找了下面两种能通过oj的解法，大概看了看，也是看的个一知半解，貌似是用两个变量 readpos 和 writepos 来记录读取和写的位置，i从0到n开始循环，如果此时读和写的位置相同，那么调用 read4 函数，将结果赋给 writepos，把 readpos 置零，如果 writepos 为零的话，说明 buf 中没有东西了，返回当前的坐标i。然后用内置的 buff 变量的 readpos 位置覆盖输入字符串 buf 的i位置，如果完成遍历，返回n，参见代码如下：

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];
};

下面这种方法和上面的方法基本相同，稍稍改变了些解法，使得看起来更加简洁一些：

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

类似题目：

read n characters given read4

参考资料：

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

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