给定三个正整数L ， R和K ，任务是从范围[L，R]中找到最大回文数组，以使该组中存在的最大元素和最小元素之间的差小于K。
例子：
Input: L = 50, R = 78, K = 12
Output: 2
Explanation:
All palindromic numbers from the range [50, 78] are {55, 66, 77}.
The group of palindromic numbers {55, 66} have the difference between the maximum and minimum element as 11, which is less than K( = 12).
Therefore, the size of the group is 2.
Input: L = 98, R = 112, K = 13
Output: 3
方法：可以通过使用二进制搜索解决给定的问题。请按照以下步骤解决问题：
- 初始化一个辅助数组，例如arr []，并存储在[L，R]范围内的所有回文数。
- 定义一个函数，例如search(arr，X)，以查找值小于X的最右边的索引：
- 初始化三个变量，比如低为0 ，高为(arr.size()– 1) ， ans为-1 。
- 迭代直到从低到高，然后执行以下操作：
- 计算中低至高+(高至低)/ 2 。
- 如果在索引中期值为至多X，然后更新ANS的作为中低的值作为(MID + 1)。
- 否则，将high的值更新为(mid – 1) 。
- 完成上述步骤后，返回ans的值作为结果。
- 遍历数组arr []并执行以下步骤：
- 使用函数search()查找最右边的索引，该索引小于或等于(arr [i] + K – 1)并将其存储在变量中，例如rightIndex 。
- 如果rightIndex的值不等于-1 ，则将count的值更新为count和(rightIndex – i + 1)的最大值。
- 完成上述步骤后，打印count的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to search the
// rightmost index of given number
static int search(vector list, int num)
{
int low = 0, high = list.size() - 1;
// Store the rightmost index
int ans = -1;
while (low <= high)
{
// Calulate the mid
int mid = low + (high - low) / 2;
// If given number <= num
if (list[mid] <= num)
{
// Assign ans = mid
ans = mid;
// Update low
low = mid + 1;
}
else
// Update high
high = mid - 1;
}
// return ans
return ans;
}
// Function to check if the given
// number is palindrome or not
bool isPalindrome(int n)
{
int rev = 0;
int temp = n;
// Generate reverse
// of the given number
while (n > 0)
{
rev = rev * 10 + n % 10;
n /= 10;
}
// If n is a palindrome
return rev == temp;
}
// Function to find the maximum size
// of group of palindrome numbers
// having difference between maximum
// and minimum element at most K
int countNumbers(int L, int R, int K)
{
// Stores the all the palindromic
// numbers in the range [L, R]
vector list;
// Travese over the range [L, R]
for(int i = L; i <= R; i++)
{
// If i is a palindrome
if (isPalindrome(i))
{
// Append the number
// in the list
list.push_back(i);
}
}
// Stores count of maximum
// palindromic numbers
int count = 0;
// Iterate each element in the list
for(int i = 0; i < list.size(); i++)
{
// Calulate rightmost index in
// the list < current elemnt + K
int right_index = search(list,
list[i] + K - 1);
// Check if there is rightmost
// index from the current index
if (right_index != -1)
count = max(count, right_index - i + 1);
}
// Return the count
return count;
}
// Driver Code
int main()
{
int L = 98, R = 112;
int K = 13;
cout << countNumbers(L, R, K);
}
// This code is contributed by ipg2016107 Java
// Java program for the above approach
import java.util.*;
public class Main {
// Function to find the maximum size
// of group of palindrome numbers
// having difference between maximum
// and minimum element at most K
static int countNumbers(int L, int R, int K)
{
// Stores the all the palindromic
// numbers in the range [L, R]
ArrayList list
= new ArrayList<>();
// Travese over the range [L, R]
for (int i = L; i <= R; i++) {
// If i is a palindrome
if (isPalindrome(i)) {
// Append the number
// in the list
list.add(i);
}
}
// Stores count of maximum
// palindromic numbers
int count = 0;
// Iterate each element in the list
for (int i = 0; i < list.size(); i++) {
// Calulate rightmost index in
// the list < current elemnt + K
int right_index
= search(list, list.get(i) + K - 1);
// Check if there is rightmost
// index from the current index
if (right_index != -1)
count = Math.max(count,
right_index - i + 1);
}
// Return the count
return count;
}
// Function to search the
// rightmost index of given number
static int search(
ArrayList list, int num)
{
int low = 0, high = list.size() - 1;
// Store the rightmost index
int ans = -1;
while (low <= high) {
// Calulate the mid
int mid = low + (high - low) / 2;
// If given number <= num
if (list.get(mid) <= num) {
// Assign ans = mid
ans = mid;
// Update low
low = mid + 1;
}
else
// Update high
high = mid - 1;
}
// return ans
return ans;
}
// Function to check if the given
// number is palindrome or not
static boolean isPalindrome(int n)
{
int rev = 0;
int temp = n;
// Generate reverse
// of the given number
while (n > 0) {
rev = rev * 10 + n % 10;
n /= 10;
}
// If n is a palindrome
return rev == temp;
}
// Driver Code
public static void main(String args[])
{
int L = 98, R = 112;
int K = 13;
System.out.print(
countNumbers(L, R, K));
}
} Python3
# Python3 program for the above approach
# Function to find the maximum size
# of group of palindrome numbers
# having difference between maximum
# and minimum element at most K
def countNumbers(L, R, K):
# Stores the all the palindromic
# numbers in the range [L, R]
list = []
# Travese over the range [L, R]
for i in range(L, R + 1):
# If i is a palindrome
if (isPalindrome(i)):
# Append the number
# in the list
list.append(i)
# Stores count of maximum
# palindromic numbers
count = 0
# Iterate each element in the list
for i in range(len(list)):
# Calulate rightmost index in
# the list < current elemnt + K
right_index = search(list, list[i] + K - 1)
# Check if there is rightmost
# index from the current index
if (right_index != -1):
count = max(count, right_index - i + 1)
# Return the count
return count
# Function to search the
# rightmost index of given number
def search(list, num):
low, high = 0, len(list) - 1
# Store the rightmost index
ans = -1
while (low <= high):
# Calulate the mid
mid = low + (high - low) // 2
# If given number <= num
if (list[mid] <= num):
# Assign ans = mid
ans = mid
# Update low
low = mid + 1
else:
# Update high
high = mid - 1
# return ans
return ans
# Function to check if the given
# number is palindrome or not
def isPalindrome(n):
rev = 0
temp = n
# Generate reverse
# of the given number
while (n > 0):
rev = rev * 10 + n % 10
n //= 10
# If n is a palindrome
return rev == temp
# Driver Code
if __name__ == '__main__':
L, R = 98, 112
K = 13
print(countNumbers(L, R, K))
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum size
// of group of palindrome numbers
// having difference between maximum
// and minimum element at most K
static int countNumbers(int L, int R, int K)
{
// Stores the all the palindromic
// numbers in the range [L, R]
List list = new List();
// Travese over the range [L, R]
for(int i = L; i <= R; i++)
{
// If i is a palindrome
if (isPalindrome(i))
{
// Append the number
// in the list
list.Add(i);
}
}
// Stores count of maximum
// palindromic numbers
int count = 0;
// Iterate each element in the list
for(int i = 0; i < list.Count; i++)
{
// Calulate rightmost index in
// the list < current elemnt + K
int right_index = search(list,
list[i] + K - 1);
// Check if there is rightmost
// index from the current index
if (right_index != -1)
count = Math.Max(count,
right_index - i + 1);
}
// Return the count
return count;
}
// Function to search the
// rightmost index of given number
static int search(List list, int num)
{
int low = 0, high = list.Count - 1;
// Store the rightmost index
int ans = -1;
while (low <= high)
{
// Calulate the mid
int mid = low + (high - low) / 2;
// If given number <= num
if (list[mid] <= num)
{
// Assign ans = mid
ans = mid;
// Update low
low = mid + 1;
}
else
// Update high
high = mid - 1;
}
// return ans
return ans;
}
// Function to check if the given
// number is palindrome or not
static bool isPalindrome(int n)
{
int rev = 0;
int temp = n;
// Generate reverse
// of the given number
while (n > 0)
{
rev = rev * 10 + n % 10;
n /= 10;
}
// If n is a palindrome
return rev == temp;
}
// Driver Code
public static void Main(string[] args)
{
int L = 98, R = 112;
int K = 13;
Console.WriteLine(countNumbers(L, R, K));
}
}
// This code is contributed by avijitmondal1998 输出：
3时间复杂度： O((R – L)* log(R – L))
辅助空间： O(R – L)