给定正整数N ，任务是找到小于N的最大回文数，并且可以表示为两个3位数的乘积。
例子：
Input: N = 101110
Output: 101101
Explanation: The number 101101 ( = 143 * 707) is the largest palindromic number possible satisfying the conditions.
Input: N = 800000
Output: 793397
Explanation: The number 793397 ( = 869 × 913) is the largest palindromic number possible satisfying the conditions.
天真的方法：最简单的方法是生成[100，999]范围内的所有可能对，并为每对检查其乘积是否是回文，是否小于N。打印所有获得的所有此类产品的最大数量，作为必需的答案。
时间复杂度： O(N * 900 2 )
辅助空间： O(1)
替代方法：也可以基于11的所有倍数都是回文的现象来解决该问题。因此，迭代[100，999]的范围内，超过11的范围在[121，999]的倍数的范围内的每一个值，迭代，并检查在每个迭代中所要求的条件。请按照以下步骤解决问题：
- 初始化一个变量，例如num ，以存储满足给定条件的最大回文数。
- 使用变量i遍历[100，999]范围，并执行以下步骤：
- 使用变量j在11的倍数中进行遍历[121，999]范围。
- 将i和j的乘积存储在字符串x中。
- 如果X的值小于N且X是回文，则如果X> num ，则更新num的值。
- 否则，请继续迭代下一对整数。
- 使用变量j在11的倍数中进行遍历[121，999]范围。
- 完成上述步骤后，输出num的值。
下面是上述方法的实现
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the largest palindrome
// not exceeding N which can be expressed
// as the product of two 3-digit numbers
void palindrome_prod(string N){
// Stores all palindromes
vector palindrome_list;
for (int i = 101; i < 1000; i++)
{
for (int j = 121; j < 1000;
j += (i % 11 == 0) ? 1 : 11)
{
// Stores the product
int n = i * j;
string x = to_string(n);
string y = x;
reverse(y.begin(), y.end());
// Check if X is palindrome
if (x == y){
// Check n is less than N
if (n < stoi(N)){
// If true, append it
// in the list
palindrome_list.push_back(i * j);
}
}
}
}
// Print the largest palindrome
cout << (*max_element(palindrome_list.begin(),
palindrome_list.end()));
}
// Driver Code
int main()
{
string N = "101110";
// Function Call
palindrome_prod(N);
return 0;
}
// This code is contributed by mohit kumar 29 Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the largest palindrome
// not exceeding N which can be expressed
// as the product of two 3-digit numbers
static void palindrome_prod(String N){
// Stores all palindromes
Vector palindrome_list = new Vector();
for (int i = 101; i < 1000; i++)
{
for (int j = 121; j < 1000;
j += (i % 11 == 0) ? 1 : 11)
{
// Stores the product
int n = i * j;
String x = String.valueOf(n);
String y = x;
reverse(y);
// Check if X is palindrome
if (x == y){
// Check n is less than N
if (n < Integer.valueOf(N)){
// If true, append it
// in the list
palindrome_list.add(i * j);
}
}
}
}
// Print the largest palindrome
System.out.print(Collections.max(palindrome_list));
}
static String reverse(String input)
{
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--)
{
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver Code
public static void main(String[] args)
{
String N = "101110";
// Function Call
palindrome_prod(N);
}
}
// This code is contributed by 29AjayKumar Python3
# Python program for the above approach
# Function to find the largest palindrome
# not exceeding N which can be expressed
# as the product of two 3-digit numbers
def palindrome_prod(N):
# Stores all palindromes
palindrome_list = []
for i in range(101, 1000):
for j in range(121, 1000, (1 if i % 11 == 0 else 11)):
# Stores the product
n = i * j
x = str(n)
# Check if X is palindrome
if x == x[::-1]:
# Check n is less than N
if n < N:
# If true, append it
# in the list
palindrome_list.append(i * j)
# Print the largest palindrome
print(max(palindrome_list))
# Driver Code
N = 101110
# Function Call
palindrome_prod(N)C#
// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG
{
// Function to find the largest palindrome
// not exceeding N which can be expressed
// as the product of two 3-digit numbers
static void palindrome_prod(String N){
// Stores all palindromes
List palindrome_list = new List();
for (int i = 101; i < 1000; i++)
{
for (int j = 121; j < 1000;
j += (i % 11 == 0) ? 1 : 11)
{
// Stores the product
int n = i * j;
String x = String.Join("", n);
String y = x;
reverse(y);
// Check if X is palindrome
if (x == y)
{
// Check n is less than N
if (n < Int32.Parse(N))
{
// If true, append it
// in the list
palindrome_list.Add(i * j);
}
}
}
}
// Print the largest palindrome
Console.Write(palindrome_list.Max());
}
static String reverse(String input)
{
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for (l = 0; l < r; l++, r--)
{
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("", a);
}
// Driver Code
public static void Main(String[] args)
{
String N = "101110";
// Function Call
palindrome_prod(N);
}
}
// This code is contributed by 29AjayKumar 输出：
101101时间复杂度： O(N * 900 2 )
辅助空间： O(1)