我正在尝试使用 strstr 函数来计算字符串“TT”出现在 DNA 序列 ATGCTAGTATTTGGATAGATAGATAGATAGATAGATAGATAAAAAAATTTTTTTT 的次数,而不计算任何“T”的两次。它应该带有 5 个 'TT' 实例,但我的函数给了我 9 个,如果你重叠了 'TT',你会得到。我该如何解决这个问题,以便只计算“TT”的每个单独实例而没有 T 被计算两次?这是我的程序:
/***************************************************************************************/
#include <iostream>
#include <cstring>
#include <iomanip>
using namespace std;
//FUNCTION PROTOTYPES
int overlap(char *ptr1, char *ptr2);
int main()
{
//Declare and initialize objects
int count(0); // For DNA sequence
//DNA SEQUENCE
char DNA_sequence[] = "ATGCTAGTATTTGGATAGATAGATAGATAGATAGATAGATAAAAAAATTTTTTTT";
char thymine_group[] = "TT";
char *ptr1(DNA_sequence), *ptr2(thymine_group);
//Send QUOTE to function
count = overlap(ptr1, ptr2);
//Print number of occurences.
cout << "'TT' appears in DNA sequence " << count << " times" << endl;
return 0;
}
//FUNCTION 1 USING CHAR ARRAYS AND POINTERS
int overlap(char *ptr1, char *ptr2)
{
int count(0);
//Count number of occurences of strg2 in strg1.
//While function strstr does not return NULL
//increment count and move ptr1 to next section
//of strg1.
while ((ptr1=strstr(ptr1,ptr2)) != NULL)
{
count++;
ptr1++;
}
return count;
}
/**************************************************************************************************/