I\'m currently writing a quick solution for Euler Problem #4 where one must find the largest palindromic number from the product of two 3-digit numbers.
To identify if a
A you want to compare a number with its reverse it may be faster to reverse the number using division rather than converting it to a string. I still need to test the speed of it.
private static int Reverse(int num) {
int res = 0;
while (num > 0) {
int rm ;
num = Math.DivRem(num, 10, out rm);
res = res * 10 + rm;
}
return res;
}
EDIT: DivRem was about 1% faster than division and module in my computer. A speed optimization is exit if the last digit is 0:
private static int Reverse(int num) {
int res = 0;
int rm;
num = Math.DivRem(num, 10, out rm);
//Some magic value or return false, see below.
if (rm == 0) return -1 ;
res = res * 10 + rm;
while (num > 0) {
num = Math.DivRem(num, 10, out rm);
res = res * 10 + rm;
}
return res ;
}
Making the method return a bool was slightly slower than comparing to a bool in a loop in my computer, but I don't understand why. Please test in your computer.
Multiplication and bit-shifing should be faster than division but probably are not precise enough. EDIT: using long seems be precise enough.
private static int FastReverse(int num) {
int res = 0;
int q = (int)((214748365L * num) >> 31);
int rm = num - 10 * q;
num = q;
if (rm == 0) return -1;
res = res * 10 + rm;
while (num > 0) {
q = (int)((214748365L * num) >> 31);
rm = num - 10 * q;
num = q;
res = res * 10 + rm;
}
return res;
}
(214748365L * num) >> 31 is equal to i / 10 until 1,073,741,829 where 1 / 10 gives 107374182 and the multiplication + binary shifting gives 107374183.
Wouldn't reversing the number be faster?
// unchecked code, don't kill me if it doesn't even compile.
ulong Reverse(ulong number) {
ulong result = 0;
while (number > 0) {
ulong digit = number % 10;
result = result * 10 + digit;
number /= 10;
}
return result;
}
I think it might be faster to do the comparison in-place. If you reverse the string, you've got to:
If you perform the comparison in place, you do only the last step. An even then, your comparison is only half the string (or half - 0.5, in the event of an odd number of characters). Something like the following should work:
static bool IsPalindromic(string s){
int len = s.Length;
int half = len-- >> 1;
for(int i = 0; i < half; i++)
if(s[i] != s[len - i])
return false;
return true;
}
EDIT:
Although this answers the OP's question, the solutions offered by ggf31416 and configurator solve the OP's real need about 30% faster, by my tests. configurator's solution is a tiny bit faster than ggf31416's, if you convert it to a static method and use int
s instead of ulong
s (but much slower, otherwise).
Incidentally, running through these examples to solve the problem the OP mentions (finding the largest palindromic product of any two three-digit numbers) with the simple (perhaps naïve) loop below:
for(int i = 100; i < 1000; i++)
for(int j = i; j < 1000; j++) // calculations where j < i would be redundant
...
yields the following results on my machine:
IsPalindromic(product.ToString()) took 0.3064174 seconds. ggf31416Reverse(product) == product took 0.1933994 seconds. configuratorReverse(product) == product took 0.1872061 seconds.
Each produces the correct result of 913 * 993 = 906609
.
try this too: http://weblogs.sqlteam.com/mladenp/archive/2006/03/19/9350.aspx
string Reverse(string s)
{
return new string(s.ToCharArray().Reverse().ToArray());
}