最接近回文数（绝对差 Is min）

给定一个数 N。我们的任务是找到与给定数的绝对差最小且绝对差必须大于 0 的最接近回文数。

例子：

Input :  N = 121 
Output : 131 or 111   
Both having equal absolute difference
with the given number.

Input : N = 1234
Output : 1221

问：亚马逊

简单的解决方案是找到小于给定数字的最大回文数，并找到第一个大于给定数的回文数。我们可以通过简单地在给定数中减少和增加一个来找到回文数，直到我们找到这些回文数数字。

以下是上述想法的实现：

C++

// C++ Program to find the closest Palindrome
// number
#include 
using namespace std;
 
// function check Palindrome
bool isPalindrome(string n)
{
    for (int i = 0; i < n.size() / 2; i++)
        if (n[i] != n[n.size() - 1 - i])
            return false;
    return true;
}
 
// convert number into String
string convertNumIntoString(int num)
{
 
    // base case:
    if (num == 0)
        return "0";
 
    string Snum = "";
    while (num > 0) {
        Snum += (num % 10 - '0');
        num /= 10;
    }
    return Snum;
}
 
// function return closest Palindrome number
int closestPalindrome(int num)
{
 
    // case1 : largest palindrome number
    // which is smaller to given number
    int RPNum = num - 1;
 
    while (!isPalindrome(convertNumIntoString(abs(RPNum))))
        RPNum--;
 
    // Case 2 : smallest palindrome number
    // which is greater than given number
    int SPNum = num + 1;
 
    while (!isPalindrome(convertNumIntoString(SPNum)))
        SPNum++;
 
    // check absolute difference
    if (abs(num - RPNum) > abs(num - SPNum))
        return SPNum;
    else
        return RPNum;
}
 
// Driver program to test above function
int main()
{
    int num = 121;
    cout << closestPalindrome(num) << endl;
    return 0;
}

Java

// Java program to find the closest
// Palindrome number
import java.io.*;
 
class GFG{
 
// Function to check Palindrome   
public static boolean isPalindrome(String s)
{
    int left = 0;
    int right = s.length() - 1;
 
    while (left < right)
    {
        if (s.charAt(left) !=
            s.charAt(right))
        {
            return false;
        }
        left++;
        right--;
    }
    return true;
}
 
// Function return closest Palindrome number
public static void closestPalindrome(int num)
{
     
    // Case1 : largest palindrome number
    // which is smaller to given number
    int RPNum = num - 1;
     
    while (isPalindrome(Integer.toString(RPNum)) == false)
    {
        RPNum--;
    }
     
    // Case 2 : smallest palindrome number
    // which is greater than given number
    int SPNum = num + 1;
     
    while (isPalindrome(Integer.toString(SPNum)) == false)
    {
        SPNum++;
    }
 
    // Check absolute difference
    if (Math.abs(num - SPNum) <
        Math.abs(num - RPNum))
    {
        System.out.println(SPNum);
    }
    else
        System.out.println(RPNum);
}
 
// Driver code    
public static void main(String[] args)
{
    int n = 121;
     
    closestPalindrome(n);
}
}
 
// This code is contributed by kes333hav

Python3

# Python3 program to find the
# closest Palindrome number
 
# Function to check Palindrome
 
 
def isPalindrome(n):
 
    for i in range(len(n) // 2):
        if (n[i] != n[-1 - i]):
            return False
 
    return True
 
# Convert number into String
 
 
def convertNumIntoString(num):
 
    Snum = str(num)
    return Snum
 
# Function return closest Palindrome number
 
 
def closestPalindrome(num):
 
    # Case1 : largest palindrome number
    # which is smaller than given number
    RPNum = num - 1
    while (not isPalindrome(
           convertNumIntoString(abs(RPNum)))):
        RPNum -= 1
 
    # Case2 : smallest palindrome number
    # which is greater than given number
    SPNum = num + 1
    while (not isPalindrome(
           convertNumIntoString(SPNum))):
        SPNum += 1
 
    # Check absolute difference
    if (abs(num - RPNum) > abs(num - SPNum)):
        return SPNum
    else:
        return RPNum
 
 
# Driver Code
if __name__ == '__main__':
 
    num = 121
 
    print(closestPalindrome(num))
 
# This code is contributed by himanshu77

C#

// C# program to find the closest
// Palindrome number
using System;
 
class GFG{
 
// Function to check Palindrome
public static bool isPalindrome(string s)
{
    int left = 0;
    int right = s.Length - 1;
 
    while (left < right)
    {
        if (s[left] != s[right])
        {
            return false;
        }
        left++;
        right--;
    }
    return true;
}
 
// Function return closest Palindrome number
public static void closestPalindrome(int num)
{
     
    // Case1 : largest palindrome number
    // which is smaller to given number
    int RPNum = num - 1;
 
    while (isPalindrome(RPNum.ToString()) == false)
    {
        RPNum--;
    }
 
    // Case 2 : smallest palindrome number
    // which is greater than given number
    int SPNum = num + 1;
 
    while (isPalindrome(SPNum.ToString()) == false)
    {
        SPNum++;
    }
 
    // Check absolute difference
    if (Math.Abs(num - SPNum) <
        Math.Abs(num - RPNum))
    {
        Console.WriteLine(SPNum);
    }
    else
        Console.WriteLine(RPNum);
}
 
// Driver code
public static void Main(string[] args)
{
    int n = 121;
 
    closestPalindrome(n);
}
}
 
// This code is contributed by ukasp

PHP

 0)
{
    $Snum .= ($num % 10 - '0');
    $num =(int)($num / 10);
}
return $Snum;
}
 
// function return closest
// Palindrome number
function closestPalindrome($num)
{
 
// case1 : largest palindrome number
// which is smaller to given number
$RPNum = $num - 1;
 
while (!isPalindrome(convertNumIntoString(abs($RPNum))))
    $RPNum--;
 
// Case 2 : smallest palindrome number
// which is greater than given number
$SPNum = $num + 1;
 
while (!isPalindrome(convertNumIntoString($SPNum)))
    $SPNum++;
 
// check absolute difference
if (abs($num - $RPNum) > abs($num - $SPNum))
    return $SPNum;
else
    return $RPNum;
}
 
    // Driver code
    $num = 121;
    echo closestPalindrome($num)."\n";
 
// This code is contributed by mits
?>

Javascript

 
// JavaScript Program to find the closest Palindrome
// number
 
// function check Palindrome
function isPalindrome(n)
{
    for (let i = 0; i < Math.floor(n.length / 2); i++)
        if (n[i] != n[n.length - 1 - i])
            return false;
    return true;
}
 
// convert number into String
function convertNumIntoString(num)
{
 
    // base case:
    if (num == 0)
        return "0";
 
    let Snum = num + '';
    return Snum;
}
 
// function return closest Palindrome number
function closestPalindrome(num)
{
 
    // case1 : largest palindrome number
    // which is smaller to given number
    let RPNum = num - 1;
 
    while (!isPalindrome(convertNumIntoString(Math.abs(RPNum))))
        RPNum--;
 
    // Case 2 : smallest palindrome number
    // which is greater than given number
    let SPNum = num + 1;
 
    while (!isPalindrome(convertNumIntoString(SPNum)))
        SPNum++;
 
    // check absolute difference
    if (Math.abs(num - RPNum) > Math.abs(num - SPNum))
        return SPNum;
    else
        return RPNum;
}
 
// Driver program to test above function
let num = 121;
document.write(closestPalindrome(num),"</br>");
     
// This code is contributed by shinjanpatra.

输出：

111

一个有效的解决方案是考虑以下情况。
案例 1：如果一个数字包含所有 9，那么我们只需在其中添加 2 即可得到下一个最接近的回文数。 num = 999：输出：num + 2 = 1001。
案例二：
案例2 a：通过复制前半部分并在最后添加镜像来获得最接近回文的一种可能方法。左半边：例如“ 123 456”的左边是“123”，“ 12 345”的左半边是“1 2”。要转换为回文，我们可以取其左半边的镜像或取其右半边的镜像。然而，如果我们取右半边的镜像，那么这样形成的回文并不能保证是最接近的回文。所以，我们必须把左边的镜子复制到右边。

Let's number : 123456 
After copy and append reverse of it at the end number looks like:
we get palindrome 123321

案例 2 b 和 2c：另外两种可能的方法来获得最接近的回文数，方法是在回文数上将中间数字减一和加一。

下面是上述想法的实现：

C++

// C++ program to find the closest Palindrome number
#include 
using namespace std;
 
#define CToI(x) (x - '0')
#define IToC(x) (x + '0')
 
// function check Palindrome
bool isPalindrome(string n)
{
    for (int i = 0; i < n.size() / 2; i++)
        if (n[i] != n[n.size() - 1 - i])
            return false;
    return true;
}
 
// check all 9's
bool checkAll9(string num)
{
    for (int i = 0; i < num.size(); i++)
        if (num[i] != '9')
            return false;
    return true;
}
 
// Add carry to the number of given size
string carryOperation(string num, int carry, int size)
{
    if (carry == -1)
    {
        int i = size - 1;
        while (i >= 0 && num[i] == '0')
            num[i--] = '9';
        if (i >= 0)
            num[i] = IToC(CToI(num[i]) - 1);
    }
    else
    {
        for (int i = size - 1; i >= 0; i--)
        {
            int digit = CToI(num[i]);
            num[i] = IToC((digit + carry) % 10);
            carry = (digit + carry) / 10;
        }
    }
    return num;
}
 
// function return the closest number
// to given number
string MIN(long long int num,
           long long int num1,
           long long int num2,
           long long int num3)
{
 
    long long int Diff1 = abs(num - num1);
    long long int Diff2 = abs(num - num2);
    long long int Diff3 = abs(num3 - num);
 
    if (Diff1 < Diff2 && Diff1 < Diff3 &&
        num1 != num)
        return to_string(num1);
    else if (Diff3 < Diff2 && (Diff1 == 0 ||
             Diff3 < Diff1))
        return to_string(num3);
    else
        return to_string(num2);
}
 
// function return closest Palindrome number
string closestPlandrome(string num)
{
 
    // base case
    if (num.size() == 1)
        return (to_string(stoi(num) - 1));
 
    // case 2:
    // If a number contains all 9's
    if (checkAll9(num))
    {
        string str = "1";
        return str.append(num.size() - 1, '0') + "1";
    }
 
    int size_ = num.size();
 
    // case 1 a:
    // copy first half and reverse it and append it
    // at the end of first half
    string FH = num.substr(0, size_ / 2);
    string odd;
 
    // odd length
    if (size_ % 2 != 0)
        odd = num[size_ / 2];
 
    // reverse
    string SH = FH;
    reverse(SH.begin(), SH.end());
 
    // store three nearest Palindrome numbers
    string RPNUM = "", EPNUM = "", LPNUM = "";
    string tempFH = "";
    string tempSH = "";
 
    if (size_ % 2 != 0)
    {
        EPNUM = FH + odd + SH;
        if (odd == "0")
        {
            tempFH = carryOperation(FH, -1, FH.size());
            tempSH = tempFH;
            reverse(tempSH.begin(), tempSH.end());
            RPNUM = tempFH + "9" + tempSH;
        }
        else
            RPNUM = FH + to_string(stoi(odd) - 1) + SH;
 
        // To handle carry
        if (odd == "9")
        {
            tempFH = carryOperation(FH, 1, FH.size());
            tempSH = tempFH;
            reverse(tempSH.begin(), tempSH.end());
            LPNUM = tempFH + "0" + tempSH;
        }
        else
            LPNUM = FH + to_string(stoi(odd) + 1) + SH;
    }
 
    // for even case
    else
    {
        int n = FH.size();
        tempFH = FH;
        EPNUM = FH + SH;
        if (FH[n - 1] == '0')
            tempFH = carryOperation(FH, -1, n);
        else
            tempFH[n - 1] = IToC(CToI(FH[n - 1]) - 1);
 
        tempSH = tempFH;
        reverse(tempSH.begin(), tempSH.end());
        RPNUM = tempFH + tempSH;
 
        tempFH = FH;
        if (FH[n - 1] == '9')
            tempFH = carryOperation(FH, 1, n);
        else
            tempFH[n - 1] = IToC(CToI(tempFH[n - 1]) + 1);
 
        tempSH = tempFH;
        reverse(tempSH.begin(), tempSH.end());
        LPNUM = tempFH + tempSH;
    }
 
    // return the closest palindrome numbers
    return MIN(stoll(num), stoll(EPNUM), stoll(RPNUM),
                                         stoll(LPNUM));
}
 
// Driver program to test above function
int main()
{
    string num = "123456";
    cout << closestPlandrome(num) << endl;
    return 0;
}

输出：

123321

时间复杂度：O(d)（d 是给定数字的位数）