[LeetCode] Read N Characters Given Read4 I & II
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.
1 // Forward declaration of the read4 API. 2 int read4(char *buf); 3 4 class Solution { 5 public: 6 /** 7 * @param buf Destination buffer 8 * @param n Maximum number of characters to read 9 * @return The number of characters read 10 */ 11 int read(char *buf, int n) { 12 char tmp[4]; 13 int idx = 0, cnt4; 14 while (idx < n) { 15 cnt4 = read4(tmp); 16 for (int i = 0; i < cnt4 && idx < n; ++i) { 17 buf[idx++] = tmp[i]; 18 } 19 if (cnt4 < 4) break; 20 } 21 return idx; 22 } 23 };
Read N Characters Given Read4 II - Call multiple times
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 may be called multiple times.
1 // Forward declaration of the read4 API. 2 int read4(char *buf); 3 4 class Solution { 5 private: 6 char tmp[4]; 7 int tmp_idx = 0, tmp_len = 0; 8 public: 9 /** 10 * @param buf Destination buffer 11 * @param n Maximum number of characters to read 12 * @return The number of characters read 13 */ 14 int read(char *buf, int n) { 15 int idx = 0; 16 bool flag; 17 while (idx < n) { 18 flag = true; 19 if (tmp_idx == tmp_len) { 20 tmp_idx = 0; 21 tmp_len = read4(tmp); 22 if (tmp_len != 4) flag = false; 23 } 24 for (; tmp_idx < tmp_len && idx < n; ++tmp_idx) { 25 buf[idx++] = tmp[tmp_idx]; 26 } 27 if (!flag) break; 28 } 29 return idx; 30 } 31 };
