[Leetcode] 157. Read N Characters Given Read4 解题报告

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.

思路：

题目的思路不是很难，主要是判断各个终止条件：如果还有剩余的字符需要读取（n > 0），那么就调用API函数进行读取。读取之后需要判断返回值，一旦发现返回值为0，说明文件中的内容已经被全部读取，则提前返回。否则就需要判断返回值和剩余需要读取的字符的大小，根据比较的结果向buffer中复制读取的内容，并且更新n的值。

代码：

// Forward declaration of the read4 API.int read4(char *buf);class Solution {public:    /**     * @param buf Destination buffer     * @param n   Maximum number of characters to read     * @return    The number of characters read     */    int read(char *buf, int n) {        int ret = 0;        char temp[4];        while(n > 0) {                          // still have content to read            int num = read4(temp);            if(num == 0) {                      // nothing left to read                break;            }            if(n > num) {                       // put all the content to buffer                for(int i = 0; i < num; ++i) {                    buf[ret++] = temp[i];                }                n -= num;            }            else {                              // only partial contents are left to put into the buffer                for(int i = 0; i < n; ++i) {                    buf[ret++] = temp[i];                }                n = 0;            }        }        return ret;    }};