给定字母数字字符串str ，任务是重新排列字符串，使得没有两个相邻的字符是相同类型的，即没有两个相邻的字符可以是字母或数字。如果没有这样的安排是可能的，打印-1 。
例子：
Input: str = “geeks2020”
Output: g2e0e2k0s
Input: str = “IPL20”
Output: I2P0L
朴素方法：最简单的方法是生成给定字符串的所有可能排列，对于每个排列，检查它是否满足给定条件。如果任何排列的结果都为真，则打印该排列。如果不存在这样的排列，则打印 -1 。
时间复杂度： O(2 N )
辅助空间： O(1)
高效的方法：为了优化上述方法，其思想是将所有字母和数字分开存储，并通过将它们交替放置在结果字符串重新排列它们。如果字母和数字的计数相差超过 1，则打印-1，因为不可能有所需的排列。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to rearrange given
// alphanumeric string such that
// no two adjacent characters
// are of the same type
string rearrange(string s)
{
// Stores alphabets and digits
string s1 = "", s2 = "";
// Store the alphabets and digits
// separately in the strings
for (char x : s) {
isalpha(x) ? s1.push_back(x)
: s2.push_back(x);
}
// Stores the count of
// alphabets and digits
int n = s1.size();
int m = s2.size();
// If respective counts
// differ by 1
if (abs(n - m) > 1)
// Desired arrangement
// not possible
return "-1";
// Stores the indexes
int i = 0, j = 0, k = 0;
// Check if first character
// should be alphabet or digit
int flag = (n >= m) ? 1 : 0;
// Place alphabets and digits
// alternatively
while (i < n and j < m) {
// If current character
// needs to be alphabet
if (flag)
s[k++] = s1[i++];
// If current character
// needs to be a digit
else
s[k++] = s2[j++];
// Flip flag for alternate
// arrangement
flag = !flag;
}
// Return resultant string
return s;
}
// Driver Code
int main()
{
// Given String
string str = "geeks2020";
// Function Call
cout << rearrange(str) << endl;
return 0;
} Java
// Java program to implement
// the above approach
class GFG{
// Function to rearrange given
// alphanumeric String such that
// no two adjacent characters
// are of the same type
static String rearrange(String s)
{
// Stores alphabets and digits
String s1 = "", s2 = "", ans = "";
char []s3 = s.toCharArray();
// Store the alphabets and digits
// separately in the Strings
for (char x : s3)
{
if(x >= 'a' && x <= 'z')
s1 += x ;
else
s2 += x;
}
// Stores the count of
// alphabets and digits
int n = s1.length();
int m = s2.length();
// If respective counts
// differ by 1
if (Math.abs(n - m) > 1)
// Desired arrangement
// not possible
return "-1";
// Stores the indexes
int i = 0, j = 0, k = 0;
// Check if first character
// should be alphabet or digit
int flag = (n >= m) ? 1 : 0;
// Place alphabets and digits
// alternatively
while (i < n && j < m)
{
// If current character
// needs to be alphabet
if (flag != 0)
ans += s1.charAt(i++);
// If current character
// needs to be a digit
else
ans += s2.charAt(j++);
// Flip flag for alternate
// arrangement
if(flag == 1)
flag = 0;
else
flag = 1;
}
// Return resultant String
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Given String
String str = "geeks2020";
// Function Call
System.out.print(rearrange(str) + "\n");
}
}
// This code is contributed by gauravrajput1Python3
# Python3 program to implement
# the above approach
# Function to rearrange given
# alphanumeric such that no
# two adjacent characters
# are of the same type
def rearrange(s):
# Stores alphabets and digits
s1 = []
s2 = []
# Store the alphabets and digits
# separately in the strings
for x in s:
if x.isalpha():
s1.append(x)
else:
s2.append(x)
# Stores the count of
# alphabets and digits
n = len(s1)
m = len(s2)
# If respective counts
# differ by 1
if (abs(n - m) > 1):
# Desired arrangement
# not possible
return "-1"
# Stores the indexes
i = 0
j = 0
k = 0
# Check if first character
# should be alphabet or digit
flag = 0
if (n >= m):
flag = 1
else:
flag = 0
# Place alphabets and digits
# alternatively
while (i < n and j < m):
# If current character
# needs to be alphabet
if (flag):
s[k] = s1[i]
k += 1
i += 1
# If current character
# needs to be a digit
else:
s[k] = s2[j]
k += 1
j += 1
# Flip flag for alternate
# arrangement
flag = not flag
# Return resultant string
return "".join(s)
# Driver Code
if __name__ == '__main__':
# Given String
str = "geeks2020"
str1 = [i for i in str]
# Function call
print(rearrange(str1))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to rearrange given
// alphanumeric String such that
// no two adjacent characters
// are of the same type
static String rearrange(String s)
{
// Stores alphabets and digits
String s1 = "", s2 = "", ans = "";
char []s3 = s.ToCharArray();
// Store the alphabets and digits
// separately in the Strings
foreach (char x in s3)
{
if(x >= 'a' && x <= 'z')
s1 += x ;
else
s2 += x;
}
// Stores the count of
// alphabets and digits
int n = s1.Length;
int m = s2.Length;
// If respective counts
// differ by 1
if (Math.Abs(n - m) > 1)
// Desired arrangement
// not possible
return "-1";
// Stores the indexes
int i = 0, j = 0, k = 0;
// Check if first character
// should be alphabet or digit
int flag = (n >= m) ? 1 : 0;
// Place alphabets and digits
// alternatively
while (i < n && j < m)
{
// If current character
// needs to be alphabet
if (flag != 0)
ans += s1[i++];
// If current character
// needs to be a digit
else
ans += s2[j++];
// Flip flag for alternate
// arrangement
if(flag == 1)
flag = 0;
else
flag = 1;
}
// Return resultant String
return ans;
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String str = "geeks2020";
// Function Call
Console.Write(rearrange(str) + "\n");
}
}
// This code is contributed by 29AjayKumar输出：
g2e0e2k00
时间复杂度： O(N)
辅助空间： O(N)
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