我正在尝试使用strstr函数计算字符串'TT'出现在DNA序列ATGCTAGTATTTGGATAGATAGAGAGATAGATAGATAGATAAAAAAATTTTTTTT中的次数,而不计算任何'T'两次。它应该出现5个'TT'实例,但我的功能是给我9,如果你重叠'TT',你会得到的。我该如何解决这个问题,以便只计算'TT'的每个实例并且不计算两次T?这是我的计划:
/***************************************************************************************/
#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;
}
/**************************************************************************************************/
答案 0 :(得分:7)
只需将循环中的ptr1++;
更改为ptr1 += strlen(ptr2);
答案 1 :(得分:0)
试
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 += strlen (ptr2);
}
return count;
}
答案 2 :(得分:0)
int count(char *haystack, char* needle)
{ int c = 0;
for(;*haystack;haystack++){
if(strcmp(haystack, needle)==0){
c++;
haystack+=strlen(needle)-1;
}
}
return c;
}