给定值N ，找出最大回文数，该数是两个N位数字的乘积。
例子：
Input: N = 2
Output: 9009
Explanation:
9009 is the largest number which is product of two 2-digit numbers 91 and 99 (9009 = 91*99)
Input: N = 3
Output: 906609
Input: N = 4
Output: 99000099
观察：对于上述问题，可以进行以下观察：
令N = 2，则乘积将包含4位数字。由于乘积将是回文，因此将采用“ abba ”的形式，其中a，b是其各自位置值的数字。
所以，
For N = 2:
“abba” = 1000a + 100b + 10b + a
= 1001a + 110b
= 11.(91a + 10b)
Similarly, for N = 3:
“abccba” = 100000a + 10000b + 1000c + 100c + 10b + 1a
= 100001a + 10010b + 1100c
= 11.(9091a + 910b + 100c)
For N = 5:
“abcdeedcba” = 1000000000a + 100000000b + 10000000c + 1000000d + 100000e + 10000e + 1000d + 100c + 10b + a
= 1000000001a + 100000010b + 10000100c + 100100d + 110000e
= 11.(90909091a + 909091b + 91000c + 10000d)
方法：从以上观察结果可以看出，每个回文产品总是有11因子。
- 对于任何N位数字P和Q，如果P和Q的乘积是回文，则P或Q可以被11整除，但不能被两者整除。
- 因此，代替检查所有可能的P和Q对的P和Q的乘积是否是回文，我们可以通过仅检查一个数字的11的倍数来减少计算次数。
下面是上述方法的实现：
Java
// Java implementation of the above approach
public class GFG {
// Function to check if a number is a
// Palindrome or not
static boolean isPalindrome(long x)
{
// Taking the string value
// of the number
String num = String.valueOf(x);
boolean result = true;
int i = 0;
int j = num.length() - 1;
// Loop to check if every i-th
// character from beginning is
// equal to every (N - i)th char
while (i < j && result) {
result = num.charAt(i++)
== num.charAt(j--);
}
return result;
}
// Function to find the largest palindrome
// which is a product of two N digited numbers
public static void find(final int nDigits)
{
// Find lowerBound, upperBound for
// a given nDigits. for n=2; [10, 99]
final long lowerBound
= (long)Math.pow(10, nDigits - 1);
final long upperBound
= (lowerBound * 10) - 1;
// Result variables
long resultP = 0, resultQ = 0,
resultR = 0;
// Keep p decrementing by 11
for (long p = upperBound;
p > lowerBound;
p -= 11) {
// Find the nearest number
// divisible by 11
while (p % 11 != 0) {
p--;
}
// Keep decrementing q by 1
for (long q = upperBound;
q > lowerBound;
q--) {
long t = p * q;
// Update the result if
// t > r and is a palindrome
if (t > resultR
&& isPalindrome(t)) {
resultP = p;
resultQ = q;
resultR = t;
break;
}
}
}
// Printing the final result
System.out.println(resultR);
}
// Driver code
public static void main(String[] args)
{
int N = 2;
find(N);
}
}Python3
# Python 3 implementation of
# the above approach
# Function to check if a
# number is a Palindrome
# or not
def isPalindrome(x):
# Taking the string value
# of the number
num = str(x)
result = True
i = 0
j = len(num) - 1
# Loop to check if every i-th
# character from beginning is
# equal to every(N - i)th char
while (i < j and result):
result = num[i] == num[j]
i += 1
j -= 1
return result
# Function to find the largest
# palindrome which is a product
# of two N digited numbers
def find(nDigits):
# Find lowerBound, upperBound
# for a given nDigits. for n = 2
# [10, 99]
lowerBound = pow(10, nDigits - 1)
upperBound = (lowerBound * 10) - 1
# Result variables
resultP = 0
resultQ = 0
resultR = 0
# Keep p decrementing by 11
for p in range(upperBound,
lowerBound, -11):
# Find the nearest number
# divisible by 11
while (p % 11 != 0):
p -= 1
# Keep decrementing q by 1
for q in range(upperBound,
lowerBound, -1):
t = p * q
# Update the result if
# t > r and is a palindrome
if (t > resultR and
isPalindrome(t)):
resultP = p
resultQ = q
resultR = t
break
# Printing the final result
print(resultR)
# Driver code
if __name__ == "__main__":
N = 2
find(N)
# This code is contributed by ChitranayalC#
// C# implementation of the above approach
using System;
class GFG {
// Function to check if a number is a
// Palindrome or not
static bool isPalindrome(long x)
{
// Taking the string value
// of the number
String num = String.Join("",x);
bool result = true;
int i = 0;
int j = num.Length - 1;
// Loop to check if every i-th
// character from beginning is
// equal to every (N - i)th char
while (i < j && result) {
result = num[i++]
== num[j--];
}
return result;
}
// Function to find the largest palindrome
// which is a product of two N digited numbers
public static void find(int nDigits)
{
// Find lowerBound, upperBound for
// a given nDigits. for n=2; [10, 99]
long lowerBound
= (long)Math.Pow(10, nDigits - 1);
long upperBound
= (lowerBound * 10) - 1;
// Result variables
long resultP = 0, resultQ = 0,
resultR = 0;
// Keep p decrementing by 11
for (long p = upperBound;
p > lowerBound;
p -= 11) {
// Find the nearest number
// divisible by 11
while (p % 11 != 0) {
p--;
}
// Keep decrementing q by 1
for (long q = upperBound;
q > lowerBound;
q--) {
long t = p * q;
// Update the result if
// t > r and is a palindrome
if (t > resultR
&& isPalindrome(t)) {
resultP = p;
resultQ = q;
resultR = t;
break;
}
}
}
// Printing the readonly result
Console.WriteLine(resultR);
}
// Driver code
public static void Main(String[] args)
{
int N = 2;
find(N);
}
}
// This code is contributed by sapnasingh4991输出：
9009
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