简单算法复杂度计算

void reverse( string &s ){
    int n = s.size();
    for(int i = 0 ; i < n/2 ; i++ )//这个循环走了n/2次，所以O(n)
        swap( s[i] , s[n-1-i] );
}

int count = 1;      
while (count < n){//cout*2^x=n,x=logn,所以O(logn)
   count = count * 2;
}

int i, j;      
for(i = 1; i < n; i++)//n^3
    for(j = 1; j < n; j++)
        for(j = 1; j < n; j++){
               ......
        }

int i, j;      
for(i = 0; i < n; i++){//n^2
    for(j = i; j < n; j++){   
        .....
    }
}

for(int i=2;i*i<=n;i++){//sqrt(n)
    ....
}