给定一个数组，该数组是原始算术级数，但其中一个元素已更改。任务是使其再次成为算术级数。如果有很多这样的可能序列，请返回其中任何一个。数组的长度将始终大于2。
例子：
Input : arr = [1, 3, 4, 7]
Output : arr = [1, 3, 5, 7]
The common difference for the arithmetic progression is 2, so the resulting
arithmetic progression is [1, 3, 5, 7].
Input : arr = [1, 3, 7]
Output : arr = [-1, 3, 7]
The common difference can be either 2 or 3 or 4.
Hence the resulting array can be either [1, 3, 5], [-1, 3, 7] or [1, 4, 7].
As any one can be chosen, [-1, 3, 7] is returned as the output.
方法：
算术级数的关键组成部分是初始项和共同点。因此，我们的任务是找到这两个值，以了解实际的算术级数。
- 如果数组的长度为3，则将公共差作为任意两个元素之间的差。
- 否则，请尝试使用前四个元素找到共同的区别。
- 使用公式a [1] – a [0] = a [2] – a [1]检查前三个元素是否处于算术级数中。如果它们在算术级数上，则公共差d = a [1] – a [0]，并且初始项将是a [0]
- 使用公式a [2] – a [1] = a [3] – a [2]检查第二，第三和第四元素是否处于算术级数。如果它们在算术级数上，则公共差d = a [2] – a [1]，这意味着第一个元素已更改，因此初始项为a [0] = a [1] – d
- 在以上两种情况下，我们都检查了第一个元素或第四个元素是否已更改。如果两种情况都为假，则表示第一个或第四个元素未更改。因此，可以将共同差取为d =(a [3] – a [0])/ 3，而初始项= a [0]
- 使用初始术语和共同差异打印所有元素。
下面是上述方法的实现：
C/C++
// C++ program to change one element of an array such
// that the resulting array is in arithmetic progression.
#include
using namespace std;
// Finds the initial term and common difference and
// prints the resulting sequence.
void makeAP(int arr[], int n)
{
int initial_term, common_difference;
if (n == 3) {
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else if ((arr[1] - arr[0]) == arr[2] - arr[1]) {
// Check if the first three elements are in
// arithmetic progression
initial_term = arr[0];
common_difference = arr[1] - arr[0];
}
else if ((arr[2] - arr[1]) == (arr[3] - arr[2])) {
// Check if the first element is not
// in arithmetic progression
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else {
// The first and fourth element are
// in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3;
initial_term = arr[0];
}
// Print the arithmetic progression
for (int i = 0; i < n; i++)
cout << initial_term + (i * common_difference) << " ";
cout << endl;
}
// Driver Program
int main()
{
int arr[] = { 1, 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
makeAP(arr, n);
return 0;
} Java
// Java program to change one element of an array such
// that the resulting array is in arithmetic progression.
import java.util.Arrays;
class AP {
static void makeAP(int arr[], int n)
{
int initial_term, common_difference;
// Finds the initial term and common difference and
// prints the resulting array.
if (n == 3) {
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else if ((arr[1] - arr[0]) == arr[2] - arr[1]) {
// Check if the first three elements are in
// arithmetic progression
initial_term = arr[0];
common_difference = arr[1] - arr[0];
}
else if ((arr[2] - arr[1]) == (arr[3] - arr[2])) {
// Check if the first element is not
// in arithmetic progression
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else {
// The first and fourth element are
// in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3;
initial_term = arr[0];
}
// Print the arithmetic progression
for (int i = 0; i < n; i++)
System.out.print(initial_term +
(i * common_difference) + " ");
System.out.println();
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 3, 7 };
int n = arr.length;
makeAP(arr, n);
}
}Python3
# Python program to change one element of an array such
# that the resulting array is in arithmetic progression.
def makeAP(arr, n):
initial_term, common_difference = 0, 0
if (n == 3):
common_difference = arr[2] - arr[1]
initial_term = arr[1] - common_difference
elif((arr[1] - arr[0]) == arr[2] - arr[1]):
# Check if the first three elements are in
# arithmetic progression
initial_term = arr[0]
common_difference = arr[1] - arr[0]
elif((arr[2] - arr[1]) == (arr[3] - arr[2])):
# Check if the first element is not
# in arithmetic progression
common_difference = arr[2] - arr[1]
initial_term = arr[1] - common_difference
else:
# The first and fourth element are
# in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3
initial_term = arr[0]
# Print the arithmetic progression
for i in range(n):
print(int(initial_term+
(i * common_difference)), end = " ")
print()
# Driver code
arr = [1, 3, 7]
n = len(arr)
makeAP(arr, n)C#
// C# program to change one element of an array such
// that the resulting array is in arithmetic progression.
using System;
public class AP
{
static void makeAP(int []arr, int n)
{
int initial_term, common_difference;
// Finds the initial term and common difference and
// prints the resulting array.
if (n == 3)
{
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else if ((arr[1] - arr[0]) == arr[2] - arr[1])
{
// Check if the first three elements are in
// arithmetic progression
initial_term = arr[0];
common_difference = arr[1] - arr[0];
}
else if ((arr[2] - arr[1]) == (arr[3] - arr[2]))
{
// Check if the first element is not
// in arithmetic progression
common_difference = arr[2] - arr[1];
initial_term = arr[1] - common_difference;
}
else
{
// The first and fourth element are
// in arithmetic progression
common_difference = (arr[3] - arr[0]) / 3;
initial_term = arr[0];
}
// Print the arithmetic progression
for (int i = 0; i < n; i++)
Console.Write(initial_term +
(i * common_difference) + " ");
Console.WriteLine();
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 3, 7 };
int n = arr.Length;
makeAP(arr, n);
}
}
// This code contributed by Rajput-Ji输出：
-1 3 7
时间复杂度：O(N)，其中N是数组中元素的数量。