蛮力法解决字符串匹配问题
以下代码使用蛮力法实现字符串S和T匹配,并统计了代码运行时间。
#include <iostream>
#include <stdio.h>
#include <windows.h>
#include <iomanip>
using namespace std;
int BF(char S[], char T[]){
int i=0,j=0;
while((S[i] != '\0') && (T[j] != '\0')){
if(S[i] == T[j]){
i++;
j++;
}
else {
i=i-j+1;
j=0;
}
}
if(T[j] == '\0')
return (i-j+1);
else
return 0;
}
int main()
{
LARGE_INTEGER nFreq;
LARGE_INTEGER nBeginTime;
LARGE_INTEGER nEndTime;
double time;
QueryPerformanceFrequency(&nFreq);
double sum = 0;
int bf;
for(int i=0; i<200; i++){
QueryPerformanceCounter(&nBeginTime);
char S[] = "ababcabccabcacbab";
char T[] = "abcac";
bf = BF(S,T);
QueryPerformanceCounter(&nEndTime);
time = (double)(nEndTime.QuadPart-nBeginTime.QuadPart)*1000000000/(double)(nFreq.QuadPart);
sum = sum+time;
}
double avgTime = sum/200.0;
cout << "Result: " << bf << endl;
cout << "The average time of BF is " << avgTime;
return 0;
}