给定长度为N的数组arr []和另一个包含{a 1 ，a 2 ，…a k }的数组P [] ，该数组代表给定数组arr []的位置，任务是检查该数组是否可以排序通过仅交换元素的arr [a i ]，arr [a i + 1 ]，其中“ i”是数组P []中的某个元素。
例子：
Input: arr[] = {3, 2, 1}, P[] = {1, 2}
Output: Yes
Explanation:
Initially, i = 1 (i.e.) first element and second element are swapped. Therefore, arr[0] <=> arr[1]. arr[] = {2, 3, 1}.
Similarly, i = 2 (i.e.) second element and third element are swapped. arr[] = {2, 1, 3}.
Finally, i = 1 (i.e.) first element and second element are swapped. arr[] = {1, 2, 3}.
Since this array is sorted, therefore, the given array can be sorted.
Input: arr[] = {5, 3, -4, 1, 12}, P[] = {2, 4, 3}
Output: No
方法：想法是使用两个指针方法来检查数组是否可以排序。
- 最初，我们创建一个大小为N的位置数组pos [] 。该数组将用于标记数组P []中的给定位置。那是：
if j = ai (1 ≤ i ≤ K)
then the element pos[ai-1] will be 1
else 0
- 现在，遍历数组并检查pos [i] = 1 。
- 如果遇到pos [i] = 1 ，我们将迭代器存储在一个临时变量中，然后将迭代器的值增加1 ，直到我们连续拥有pos [i] = 1 ，即，
j = i
while (j < N and pos[j])
j=j+1
- 在此增量之后，我们将从i获得的该段排序为j + 1，最后，在必须检查的向量中的位置j之后进行检查，因为我们一直排序到该段。
Sort(arr[i] to arr[j+1])
i=j
- 最后，在完成循环之后，我们必须检查数组是否已排序。
下面是上述方法的实现：
C++
// C++ program to check if the array
// can be sorted only if the elements
// on the given positions can be swapped
#include
using namespace std;
// Function to check if the array can
// be sorted only if the elements on
// the given positions can be swapped
void check_vector(vector A, int n,
vector p)
{
// Creating an array for marking
// the positions
vector pos(A.size());
// Iterating through the array and
// mark the positions
for (int i = 0; i < p.size(); i++) {
pos[p[i] - 1] = 1;
}
int flag = 1;
// Iterating through the given array
for (int i = 0; i < n; i++) {
if (pos[i] == 0)
continue;
int j = i;
// If pos[i] is 1, then incrementing
// till 1 is continuously present in pos
while (j < n && pos[j])
++j;
// Sorting the required segment
sort(A.begin() + i, A.begin() + j + 1);
i = j;
}
// Checking if the vector is sorted or not
for (int i = 0; i < n - 1; i++) {
if (A[i] > A[i + 1]) {
flag = 0;
break;
}
}
// Print yes if it is sorted
if (flag == 1)
cout << "Yes";
else
cout << "No";
}
// Driver code
int main()
{
vector A{ 3, 2, 1 };
vector p{ 1, 2 };
check_vector(A, A.size(), p);
return 0;
} Java
// Java program to check if the array
// can be sorted only if the elements
// on the given positions can be swapped
import java.util.Arrays;
class GFG{
// Function to check if the array can
// be sorted only if the elements on
// the given positions can be swapped
public static void check_vector(int[] A,
int n,
int[] p)
{
// Creating an array for marking
// the positions
int[] pos = new int[A.length];
// Iterating through the array and
// mark the positions
for(int i = 0; i < p.length; i++)
{
pos[p[i] - 1] = 1;
}
int flag = 1;
// Iterating through the given array
for(int i = 0; i < n; i++)
{
if (pos[i] == 0)
continue;
int j = i;
// If pos[i] is 1, then incrementing
// till 1 is continuously present in pos
while (j < n && pos[j] != 0)
++j;
// Sorting the required segment
Arrays.sort(A, i, j + 1);
i = j;
}
// Checking if the vector is sorted or not
for(int i = 0; i < n - 1; i++)
{
if (A[i] > A[i + 1])
{
flag = 0;
break;
}
}
// Print yes if it is sorted
if (flag == 1)
System.out.print("Yes");
else
System.out.print("No");
}
// Driver code
public static void main(String[] args)
{
int A[] = { 3, 2, 1 };
int p[] = { 1, 2 };
check_vector(A, A.length, p);
}
}
// This code is contributed by divyeshrabadiya07Python3
# Python3 program to check if the array
# can be sorted only if the elements
# on the given positions can be swapped
# Function to check if the array can
# be sorted only if the elements on
# the given positions can be swapped
def check_vector(A, n, p):
# Creating an array for marking
# the positions
pos = [0 for i in range(len(A))]
# Iterating through the array and
# mark the positions
for i in range(len(p)):
pos[p[i] - 1] = 1
flag = 1
# Iterating through the given array
for i in range(n):
if (pos[i] == 0):
continue
j = i
# If pos[i] is 1, then incrementing
# till 1 is continuously present in pos
while (j < n and pos[j]):
j += 1
# Sorting the required segment
p = A[: i]
q = A[i : i + j + 1]
r = A[i + j + 1 : len(A)]
q.sort(reverse = False)
A = p + q + r
i = j
# Checking if the vector is sorted or not
for i in range(n - 1):
if (A[i] > A[i + 1]):
flag = 0
break
# Print yes if it is sorted
if (flag == 1):
print("Yes")
else:
print("No");
# Driver code
if __name__ == '__main__':
A = [ 3, 2, 1 ]
p = [ 1, 2 ]
check_vector(A,len(A), p)
# This code is contributed by SamarthC#
// C# program to check
// if the array can be
// sorted only if the
// elements on the given
// positions can be swapped
using System;
class GFG{
// Function to check if the array can
// be sorted only if the elements on
// the given positions can be swapped
public static void check_vector(int[] A,
int n,
int[] p)
{
// Creating an array for marking
// the positions
int[] pos = new int[A.Length];
// Iterating through the array and
// mark the positions
for(int i = 0; i < p.Length; i++)
{
pos[p[i] - 1] = 1;
}
int flag = 1;
// Iterating through the given array
for(int i = 0; i < n; i++)
{
if (pos[i] == 0)
continue;
int j = i;
// If pos[i] is 1, then
// incrementing till 1
// is continuously present in pos
while (j < n && pos[j] != 0)
++j;
// Sorting the required segment
Array.Sort(A, i, j + 1);
i = j;
}
// Checking if the vector
// is sorted or not
for(int i = 0; i < n - 1; i++)
{
if (A[i] > A[i + 1])
{
flag = 0;
break;
}
}
// Print yes if it is sorted
if (flag == 1)
Console.Write("Yes");
else
Console.Write("No");
}
// Driver code
public static void Main()
{
int[] A = {3, 2, 1};
int[] p = {1, 2};
check_vector(A, A.Length, p);
}
}
// This code is contributed by Chitranayal输出：
Yes
时间复杂度： O(N * log(N)) ，其中N是数组的大小。