使用最多两次交换形成最大回文数

给定一个包含n位数的非负回文数num 。问题是对数字num最多应用两次交换操作，以便结果是最大可能的回文数。
例子：

Input  : 4697557964
Output : 9647557469
In, 4697557964 the highlighted digits were
swapped to get the largest palindromic number 
9647557469.

Input : 54345
Output : 54345
No swapping of digits required.

方法：如果 n < 3，则num本身就是最大可能的回文数。否则计算mid = (n / 2) – 1。然后创建一个大小为(mid + 1)的数组rightMax[] 。 rightMax[i]包含位于num[i]右侧且大于num[i]且 0 <= i <= mid 的最大数字的索引。如果不存在这样的数字，则rightMax[i] = -1。现在，从 i = 0 到 m 遍历rightMax[]数组，找到第一个具有 rightMax[i] != -1 的元素。执行swap(num[i], num[rightMax[i]])和swap(num[n – i – 1], num[n – rightMax[i] – 1])操作并中断。

C++

// C++ implementation to form the largest palindromic
// number using atmost two swaps
#include 
 
using namespace std;
 
// function to form the largest palindromic
// number using atmost two swaps
void largestPalin(char num[], int n)
{
    // if length of number is less than '3'
    // then no higher palindromic number
    // can be formed
    if (n <= 3)
        return;
 
    // find the index of last digit
    // in the 1st half of 'num'
    int mid = n / 2 - 1;
 
    int rightMax[mid + 1], right;
 
    // as only the first half of 'num[]' is
    // being considered, therefore
    // for the rightmost digit in the first half
    // of 'num[]', there will be no greater right digit
    rightMax[mid] = -1;
 
    // index of the greatest right digit till the
    // current index from the right direction
    right = mid;
 
    // traverse the array from second right element
    // in the first half of 'num[]' up to the
    // left element
    for (int i = mid - 1; i >= 0; i--) {
 
        // if 'num[i]' is less than the greatest digit
        // encountered so far
        if (num[i] < num[right])
            rightMax[i] = right;
 
        else {
 
            // there is no greater right digit
            // for 'num[i]'
            rightMax[i] = -1;
 
            // update 'right' index
            right = i;
        }
    }
 
    // traverse the 'rightMax[]' array from left to right
    for (int i = 0; i <= mid; i++) {
 
        // if for the current digit, greater right digit exists
        // then swap it with its greater right digit and also
        // perform the required swap operation in the right halft
        // of 'num[]' to maintain palindromic property, then break
        if (rightMax[i] != -1) {
 
            // performing the required swap operations
            swap(num[i], num[rightMax[i]]);
            swap(num[n - i - 1], num[n - rightMax[i] - 1]);
            break;
        }
    }
}
 
// Driver program to test above
int main()
{
    char num[] = "4697557964";
    int n = strlen(num);
    largestPalin(num, n);
 
    // required largest palindromic number
    cout << "Largest Palindrome: "
         << num;
 
    return 0;
}

Java

// Java implementation to form the largest palindromic
// number using atmost two swaps
class GFG
{
 
// function to form the largest palindromic
// number using atmost two swaps
static void largestPalin(char num[], int n)
{
    // if length of number is less than '3'
    // then no higher palindromic number
    // can be formed
    if (n <= 3)
        return;
 
    // find the index of last digit
    // in the 1st half of 'num'
    int mid = n / 2 - 1;
 
    int []rightMax = new int[mid + 1];int right;
 
    // as only the first half of 'num[]' is
    // being considered, therefore
    // for the rightmost digit in the first half
    // of 'num[]', there will be no greater right digit
    rightMax[mid] = -1;
 
    // index of the greatest right digit till the
    // current index from the right direction
    right = mid;
 
    // traverse the array from second right element
    // in the first half of 'num[]' up to the
    // left element
    for (int i = mid - 1; i >= 0; i--)
    {
 
        // if 'num[i]' is less than the greatest digit
        // encountered so far
        if (num[i] < num[right])
            rightMax[i] = right;
 
        else
        {
 
            // there is no greater right digit
            // for 'num[i]'
            rightMax[i] = -1;
 
            // update 'right' index
            right = i;
        }
    }
 
    // traverse the 'rightMax[]' array from left to right
    for (int i = 0; i <= mid; i++)
    {
 
        // if for the current digit, greater right digit exists
        // then swap it with its greater right digit and also
        // perform the required swap operation in the right halft
        // of 'num[]' to maintain palindromic property, then break
        if (rightMax[i] != -1)
        {
 
            // performing the required swap operations
            swap(num,i, rightMax[i]);
            swap(num,n - i - 1, n - rightMax[i] - 1);
            break;
        }
    }
}
 
static char[] swap(char []arr, int i, int j)
{
    char temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
 
// Driver code
public static void main(String[] args)
{
    char num[] = "4697557964".toCharArray();
    int n = num.length;
    largestPalin(num, n);
 
    // required largest palindromic number
    System.out.println("Largest Palindrome: "
        + String.valueOf(num));
}
}
 
// This code has been contributed by 29AjayKumar

Python 3

# Python implementation to form the largest
# palindromic number using atmost two swaps
 
# function to form the largest palindromic
# number using atmost two swaps
def largestPalin(num, n):
 
    # if length of number is less than '3'
    # then no higher palindromic number
    # can be formed
    if n <= 3:
        return
 
    # find the index of last digit
    # in the 1st half of 'num'
    mid = n // 2 + 1
 
    rightMax = [0] * (mid + 1)
 
    # as only the first half of 'num[]' is
    # being considered, therefore
    # for the rightmost digit in the first half
    # of 'num[]', there will be no greater right digit
    rightMax[mid] = -1
 
    # index of the greatest right digit till the
    # current index from the right direction
    right = mid
 
    # traverse the array from second right element
    # in the first half of 'num[]' up to the
    # left element
    for i in range(mid-1, -1, -1):
         
        # if 'num[i]' is less than the greatest digit
        # encountered so far
        if num[i] < num[right]:
            rightMax[i] = right
 
        else:
 
            # there is no greater right digit
            # for 'num[i]'
            rightMax[i] = -1
 
            # update 'right' index
            right = i
 
    # traverse the 'rightMax[]' array from left to right
    for i in range(mid + 1):
 
        # if for the current digit, greater right digit exists
        # then swap it with its greater right digit and also
        # perform the required swap operation in the right halft
        # of 'num[]' to maintain palindromic property, then break
        if rightMax[i] != -1:
 
            # performing the required swap operations
            num[i], num[rightMax[i]] = num[rightMax[i]], num[i]
            num[n-i-1], num[n - rightMax[i] - 1] = num[n - rightMax[i] - 1], num[n - i - 1]
            break
 
 
# Driver Code
if __name__ == "__main__":
    num = "4697557964"
    n = len(num)
     
    # Required as string object do not
    # support item assignment
    num = list(num)
    largestPalin(num, n)
 
    # making string again from list
    num = ''.join(num)
    print("Largest Palindrome: ",num)
 
# This code is contributed by
# sanjeev2552

C#

// C# implementation to form the largest
// palindromic number using atmost two swaps
using System;
 
class GFG
{
 
// function to form the largest palindromic
// number using atmost two swaps
static void largestPalin(char []num, int n)
{
    // if length of number is less than '3'
    // then no higher palindromic number
    // can be formed
    if (n <= 3)
        return;
 
    // find the index of last digit
    // in the 1st half of 'num'
    int mid = n / 2 - 1;
 
    int []rightMax = new int[mid + 1]; int right;
 
    // as only the first half of 'num[]' is
    // being considered, therefore
    // for the rightmost digit in the first half
    // of 'num[]', there will be no greater right digit
    rightMax[mid] = -1;
 
    // index of the greatest right digit till the
    // current index from the right direction
    right = mid;
 
    // traverse the array from second right element
    // in the first half of 'num[]' up to the
    // left element
    for (int i = mid - 1; i >= 0; i--)
    {
 
        // if 'num[i]' is less than the greatest
        // digit encountered so far
        if (num[i] < num[right])
            rightMax[i] = right;
 
        else
        {
 
            // there is no greater right digit
            // for 'num[i]'
            rightMax[i] = -1;
 
            // update 'right' index
            right = i;
        }
    }
 
    // traverse the 'rightMax[]' array
    // from left to right
    for (int i = 0; i <= mid; i++)
    {
 
        // if for the current digit, greater right
        // digit exists then swap it with its greater
        // right digit and also perform the required
        // swap operation in the right half of 'num[]'
        // to maintain palindromic property, then break
        if (rightMax[i] != -1)
        {
 
            // performing the required swap operations
            swap(num, i, rightMax[i]);
            swap(num, n - i - 1, n - rightMax[i] - 1);
            break;
        }
    }
}
 
static char[] swap(char []arr, int i, int j)
{
    char temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
 
// Driver code
public static void Main(String[] args)
{
    char []num = "4697557964".ToCharArray();
    int n = num.Length;
    largestPalin(num, n);
 
    // required largest palindromic number
    Console.WriteLine("Largest Palindrome: " +
                        String.Join("", num));
}
}
 
// This code contributed by Rajput-Ji

Javascript

 
// JavaScript implementation to form the largest palindromic
// number using atmost two swaps
 
// function to form the largest palindromic
// number using atmost two swaps
function largestPalin(num,n)
{
    // if length of number is less than '3'
    // then no higher palindromic number
    // can be formed
    if (n <= 3)
        return;
   
    // find the index of last digit
    // in the 1st half of 'num'
    let mid = Math.floor(n / 2) - 1;
   
    let rightMax = new Array(mid + 1);
    let right;
   
    // as only the first half of 'num[]' is
    // being considered, therefore
    // for the rightmost digit in the first half
    // of 'num[]', there will be no greater right digit
    rightMax[mid] = -1;
   
    // index of the greatest right digit till the
    // current index from the right direction
    right = mid;
   
    // traverse the array from second right element
    // in the first half of 'num[]' up to the
    // left element
    for (let i = mid - 1; i >= 0; i--)
    {
   
        // if 'num[i]' is less than the greatest digit
        // encountered so far
        if (num[i] < num[right])
            rightMax[i] = right;
   
        else
        {
   
            // there is no greater right digit
            // for 'num[i]'
            rightMax[i] = -1;
   
            // update 'right' index
            right = i;
        }
    }
   
    // traverse the 'rightMax[]' array from left to right
    for (let i = 0; i <= mid; i++)
    {
   
        // if for the current digit, greater right digit exists
        // then swap it with its greater right digit and also
        // perform the required swap operation in the right halft
        // of 'num[]' to maintain palindromic property, then break
        if (rightMax[i] != -1)
        {
   
            // performing the required swap operations
            swap(num,i, rightMax[i]);
            swap(num,n - i - 1, n - rightMax[i] - 1);
            break;
        }
    }
}
 
function swap(arr,i,j)
{
    let temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
 
// Driver code
let num = "4697557964".split("");
let n = num.length;
largestPalin(num, n);
 
// required largest palindromic number
document.write("Largest Palindrome: "
        + (num).join(""));
 
// This code is contributed by rag2127
 

输出：

Largest Palindrome: 9647557469

时间复杂度： O(n)。
辅助空间： O(n)。