c++ - strstr() 函数重叠字符串搜索

Question

我正在尝试使用 strstr 函数来计算字符串“TT”在 DNA 序列 ATGCTAGTATTTGGATAGATAGATAGATAGATAGATAGATAAAAAAAATTTTTTTTT 中出现的次数，而不计算任何“T”两次。它应该带有 5 个“TT”实例，但我的函数却给了我 9 个，如果你将“TT”重叠，就会得到这个结果。我该如何解决这个问题，以便只计算“TT”的每个单独实例，而不计算 T 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;
}

/**************************************************************************************************/

Answer 1

只需将循环中的 ptr1++; 更改为 ptr1 += strlen(ptr2);