给定数字N ，任务是检查N是否为交替数字。如果N是一个交替数字，则打印“是”，否则打印“否” 。
Alternating Number is a positive integer for which, in base-10, the parity of its digits are alternates i.e., the digits in number N is followed by odd, even, odd, … or even, odd, even, … order.
例子：
Input: N = 129
Output: Yes
Explanation:
129 has digits which is in alternate odd even odd form.
Input: N = 28
Output: No
Explanation:
28 has digits which is not in alternate odd even odd form.
方法：想法是将数字转换为字符串，并检查是否有任何数字后跟相同奇偶校验的数字，则N不是交替数，否则N是交替数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if a string is
// of the form even odd even odd...
bool isEvenOddForm(string s)
{
int n = s.length();
for (int i = 0; i < n; i++) {
if (i % 2 == 0 && s[i] % 2 != 0)
return false;
if (i % 2 == 1 && s[i] % 2 != 1)
return false;
}
return true;
}
// Function to check if a string is
// of the form odd even odd even ...
bool isOddEvenForm(string s)
{
int n = s.length();
for (int i = 0; i < n; i++) {
if (i % 2 == 0 && s[i] % 2 != 1)
return false;
if (i % 2 == 1 && s[i] % 2 != 0)
return false;
}
return true;
}
// Function to check if n is an
// alternating number
bool isAlternating(int n)
{
string str = to_string(n);
return (isEvenOddForm(str)
|| isOddEvenForm(str));
}
// Driver Code
int main()
{
// Given Number N
int N = 129;
// Function Call
if (isAlternating(N))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java program for the above approach
class GFG{
// Function to check if a string is
// of the form even odd even odd...
static boolean isEvenOddForm(String s)
{
int n = s.length();
for(int i = 0; i < n; i++)
{
if (i % 2 == 0 && s.charAt(i) % 2 != 0)
return false;
if (i % 2 == 1 && s.charAt(i) % 2 != 1)
return false;
}
return true;
}
// Function to check if a string is
// of the form odd even odd even ...
static boolean isOddEvenForm(String s)
{
int n = s.length();
for(int i = 0; i < n; i++)
{
if (i % 2 == 0 && s.charAt(i) % 2 != 1)
return false;
if (i % 2 == 1 && s.charAt(i) % 2 != 0)
return false;
}
return true;
}
// Function to check if n is an
// alternating number
static boolean isAlternating(int n)
{
String str = Integer.toString(n);
return (isEvenOddForm(str) ||
isOddEvenForm(str));
}
// Driver Code
public static void main(String[] args)
{
// Given number N
int N = 129;
// Function call
if (isAlternating(N))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This Code is contributed by rock_coolPython3
# Python3 program for the above approach
# Function to check if a string is
# of the form even odd even odd...
def isEvenOddForm(s):
n = len(s)
for i in range(n):
if (i % 2 == 0 and
int(s[i]) % 2 != 0):
return False
if (i % 2 == 1 and
int(s[i]) % 2 != 1):
return False
return True
# Function to check if a string is
# of the form odd even odd even ...
def isOddEvenForm(s):
n = len(s)
for i in range(n):
if (i % 2 == 0 and
int(s[i]) % 2 != 1):
return False
if (i % 2 == 1 and
int(s[i]) % 2 != 0):
return False
return True
# Function to check if n is an
# alternating number
def isAlternating(n):
s = str(n)
return (isEvenOddForm(s) or
isOddEvenForm(s))
# Driver Code
# Given Number N
N = 129
# Function Call
if (isAlternating(N)):
print("Yes")
else:
print("No")
# This code is contributed by Vishal MauryaC#
// C# program for the above approach
using System;
class GFG{
// Function to check if a string is
// of the form even odd even odd...
static bool isEvenOddForm(String s)
{
int n = s.Length;
for(int i = 0; i < n; i++)
{
if (i % 2 == 0 && s[i] % 2 != 0)
return false;
if (i % 2 == 1 && s[i] % 2 != 1)
return false;
}
return true;
}
// Function to check if a string is
// of the form odd even odd even ...
static bool isOddEvenForm(String s)
{
int n = s.Length;
for(int i = 0; i < n; i++)
{
if (i % 2 == 0 && s[i] % 2 != 1)
return false;
if (i % 2 == 1 && s[i] % 2 != 0)
return false;
}
return true;
}
// Function to check if n is an
// alternating number
static bool isAlternating(int n)
{
String str = n.ToString();
return (isEvenOddForm(str) ||
isOddEvenForm(str));
}
// Driver Code
public static void Main(String[] args)
{
// Given number N
int N = 129;
// Function call
if (isAlternating(N))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Princi SinghJavascript
输出：
Yes时间复杂度： O(log 10 N)