给定大小为N的整数数组arr ,其中包含从1到N的不同元素。任务是检查是否可以找到数组中的位置,以便可以按顺时针方向或逆时针方向对从1到N的所有数字进行计数。
例子:
Input: arr[] = [2, 3, 4, 5, 1]
Output: YES
Explanation:
1 2
5 3
4
If counting is start at index 4
then all numbers can be counted
from 1 to N in a clockwise order.
Input: arr[] = {1, 2, 3, 5, 4]
Output: NO
Explanation:
There is no any index in array
from which given array can count
1 to N in clockwise order or
counterclockwise order.
方法:以上问题可以通过观察和分析解决。
- 在阵列中的索引可以发现仅当连续元素大于1之间的绝对差值的计数正好是1,因为仅然后,将有可能从1数到N的顺时针或逆时针顺序。
- 如果相邻元素之间的绝对差计数大于1 ,则将不可能按顺时针或逆时针顺序从1到N计数。
以下是上述方法的基本实现:
C++
// C++ program to check Clockwise or
// counterclockwise order in an array
#include
using namespace std;
bool check_order(vector arr)
{
int cnt = 0;
for (int i = 0; i < arr.size() - 1; i++) {
if (abs(arr[i + 1] - arr[i]) > 1)
cnt++;
}
// Comparing the first and last
// value of array
if (abs(arr[0] - arr[arr.size() - 1]) > 1)
cnt++;
// If the Count is greater
// than 1 then it can't be
// represented in required order
if (cnt > 1)
return false;
return true;
}
// Driver function
int main()
{
vector arr = { 2, 3, 4, 5, 1 };
if (check_order(arr))
cout << "YES";
else
cout << "NO";
return 0;
}
Java
// Java program to check clockwise or
// counterclockwise order in an array
class GFG{
static boolean check_order(int []arr)
{
int cnt = 0;
for(int i = 0; i < arr.length - 1; i++)
{
if (Math.abs(arr[i + 1] -
arr[i]) > 1)
cnt++;
}
// Comparing the first and last
// value of array
if (Math.abs(arr[0] -
arr[arr.length - 1]) > 1)
cnt++;
// If the Count is greater
// than 1 then it can't be
// represented in required order
if (cnt > 1)
return false;
return true;
}
// Driver code
public static void main(String[] args)
{
int []arr = { 2, 3, 4, 5, 1 };
if (check_order(arr))
System.out.print("YES");
else
System.out.print("NO");
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program to check clockwise or
# counterclockwise order in an array
def check_order(arr):
cnt = 0
for i in range(len(arr) - 1):
if (abs(arr[i + 1] - arr[i]) > 1):
cnt += 1
# Comparing the first and last
# value of array
if (abs(arr[0] - arr[len(arr) - 1]) > 1):
cnt += 1
# If the count is greater
# than 1 then it can't be
# represented in required order
if (cnt > 1):
return False
return True
# Driver code
arr = [ 2, 3, 4, 5, 1 ]
if (check_order(arr)):
print("YES")
else:
print("NO")
# This code is contributed by Vishal Maurya.
C#
// C# program to check clockwise or
// counterclockwise order in an array
using System;
class GFG{
static bool check_order(int []arr)
{
int cnt = 0;
for(int i = 0; i < arr.Length - 1; i++)
{
if (Math.Abs(arr[i + 1] -
arr[i]) > 1)
cnt++;
}
// Comparing the first and last
// value of array
if (Math.Abs(arr[0] -
arr[arr.Length - 1]) > 1)
cnt++;
// If the Count is greater
// than 1 then it can't be
// represented in required order
if (cnt > 1)
return false;
return true;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 3, 4, 5, 1 };
if (check_order(arr))
Console.Write("YES");
else
Console.Write("NO");
}
}
// This code is contributed by Amit Katiyar
输出:
YES
时间复杂度: O(N)
辅助空间: O(1)