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/*
 * A palindromic number reads the same both ways. The largest palindrome
 * made from the product of two 2-digit numbers is 9009 = 91 ×ばつ 99. Find the
 * largest palindrome made from the product of two 3-digit numbers.
 */
public static int get4() {
 int max = 1;
 for (int i = 100; i < 999; i++) {
 for (int j = i; j < 999; j++) {
 int n = i * j;
 int len = (int) Math.floor(Math.log10(n));
 boolean ispalindrome = true;
 for (int k = 0; k < 3; k++) {// 100^2=10^4,999^2=998001. 5-6
 // digits middle atmax 3
 int digitatk = (int) ((n % Math.pow(10, k + 1) - n
 % Math.pow(10, k)) / Math.pow(10, k));
 // eg. digit at 3 of 12345 =
 // (12345%10^4-12345%10^3)/10^3=(2345-345)/10^3=2000/10^3=2
 int l = len - k;
 int digitatl = (int) ((n % Math.pow(10, l + 1) - n
 % Math.pow(10, l)) / Math.pow(10, l));
 if (digitatl != digitatk) {
 ispalindrome = false;
 }
 }
 if (ispalindrome && n > max) {
 max = n;
 }
 }
 }
 return max;
}

Output:

Answer is 906609
Time taken 2.82342934 seconds

I am seeing that this is actually not in accord with the difficulty level said there. Firstly I cannot list all possible improvements it could have, so if you have any improvement help me. Current problems (not exhaustive):

• Time
• \${\mathcal O}(n^2)\$ where \$n=999\$ for looping only.
• Length checking method.
• Checking Palindrome.
• Digit Finding Method.