我寫以下函數查找最常見的對字符的字符串中的
//O(n^2)
void MostCommonPair(char * cArr , char * ch1 , char * ch2 , int * amount)
{
int count , max = 0;
char cCurrent , cCurrent2;
int i = 0 , j;
while(*(cArr + i + 1) != '\0')
{
cCurrent = *(cArr + i);
cCurrent2 = *(cArr + i + 1);
for(j = i , count = 0 ; *(cArr + j + 1) != '\0' ; j++)
{
if(cCurrent == *(cArr + j) && cCurrent2 == *(cArr + j + 1))
{
count++;
}
}
if(count > max)
{
*ch1 = cCurrent;
*ch2 = cCurrent2;
max = *amount = count;
}
i++;
}
}
以下輸入
「xdshahaalohalobscxbsbsbs」
CH1 = B CH 2 = S量= 4
但在我看來,這個函數效率很低,有沒有辦法只通過一次字符串或將運行大小減少到O(n)?
說明該OP正在尋找的字符具有最高計數的連續對。 – hatchet