给定的阵列ARR []由N字符串，任务是第一个排序字符串的奇数和偶数索引在降低和增加分别顺序，对于阵列的每个字符串，然后，排序修改后的数组中的字符。
例子：
Input: arr[] = {“ball”, “bat”, “boy”}
Output: albl atb byo
Explanation:
S[0]=”ball” is converted to “albl”
S[1]=”bat” is converted to “atb”
S[2]=”boy” is converted to “byo”
Sorted sequence of the modified array is: albl atb byo.
Input: arr[] = {“geeks”, “gfg”, “hello”, “world”}
Output: dwlro eohll esekg fgg
方法：给定的问题可以通过遍历字符串的给定阵列ARR []并为每个字符串首先排序字符串来解决，然后重新排列的每个字符串，使得在偶数索引字符是按升序排列，在奇数索引的字符是按降序排列.最后，按升序对新的字符串数组进行排序。请按照以下步骤解决问题：
- 遍历给定的字符串数组arr[]并对每个字符串执行以下步骤：
- 按字母顺序对字符串arr[i]进行排序。
- 初始化一个变量，比如temp为“” ，用于存储结果字符串。
- 遍历字符串ARR [I]的字符串ARR [I]的半长度和存储第i从一开始个索引和字符串的字符串温度的末尾。
- 如果字符串的大小为奇数，则将字符串arr[i]的中间字符添加到变量temp的末尾。
- 完成上述步骤后，将数组arr[i] 的第i个索引处的字符串更新为字符串temp 。
- 完成上述步骤后，将新的字符串数组按升序排序并打印字符串数组arr[] 。
下面是上述方法的实现。
C++
// C++ program for the above approach
#include
using namespace std;
// Function to sort even indices in
// ascending order and odd indices in
// descending order and sort the array
// of strings in ascending order
void sortString(string S[], int N)
{
// Traverse array of strings
for (int i = 0; i < N; i++) {
// Sort string in ascending order
sort(S[i].begin(), S[i].end());
// Length of string
int n = S[i].size();
string temp = "";
// Traverse the string
for (int j = 0; j < n / 2; j++) {
temp += S[i][j];
temp += S[i][n - j - 1];
}
// If length of string is odd
if (n & 1)
temp += S[i][n / 2];
S[i] = temp;
}
// Sort array of strings
sort(S, S + N);
// Print array of strings
for (int i = 0; i < N; i++) {
cout << S[i] << " ";
}
}
// Driver Code
int main()
{
string arr[] = { "ball", "bat", "boy" };
int N = sizeof(arr) / sizeof(arr[0]);
sortString(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
static String sortString(String inputString)
{
// Convert input string to char array
char tempArray[] = inputString.toCharArray();
// Sort tempArray
Arrays.sort(tempArray);
// Return new sorted string
return new String(tempArray);
}
// Function to sort even indices in
// ascending order and odd indices in
// descending order and sort the array
// of Strings in ascending order
static void sortString(String S[], int N)
{
// Traverse array of Strings
for(int i = 0; i < N; i++)
{
// Sort String in ascending order
S[i] = sortString(S[i]);
// Length of String
int n = S[i].length();
String temp = "";
// Traverse the String
for(int j = 0; j < n / 2; j++)
{
temp += S[i].charAt(j);
temp += S[i].charAt(n - j - 1);
}
// If length of String is odd
if (n %2== 1)
temp += S[i].charAt(n / 2);
S[i] = temp;
}
// Sort array of Strings
Arrays.sort(S);
// Print array of Strings
for(int i = 0; i < N; i++)
{
System.out.print(S[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
String arr[] = { "ball", "bat", "boy" };
int N = arr.length;
sortString(arr, N);
}
}
// This code is contributed by Amit KatiyarC#
// C# program for the above approach
using System;
public class GFG{
static String sortString(String inputString)
{
// Convert input string to char array
char []tempArray = inputString.ToCharArray();
// Sort tempArray
Array.Sort(tempArray);
// Return new sorted string
return new String(tempArray);
}
// Function to sort even indices in
// ascending order and odd indices in
// descending order and sort the array
// of Strings in ascending order
static void sortString(String []S, int N)
{
// Traverse array of Strings
for(int i = 0; i < N; i++)
{
// Sort String in ascending order
S[i] = sortString(S[i]);
// Length of String
int n = S[i].Length;
String temp = "";
// Traverse the String
for(int j = 0; j < n / 2; j++)
{
temp += S[i][j];
temp += S[i][n - j - 1];
}
// If length of String is odd
if (n %2== 1)
temp += S[i][n / 2];
S[i] = temp;
}
// Sort array of Strings
Array.Sort(S);
// Print array of Strings
for(int i = 0; i < N; i++)
{
Console.Write(S[i] + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
String []arr = { "ball", "bat", "boy" };
int N = arr.Length;
sortString(arr, N);
}
}
// This code is contributed by Amit Katiyar输出：
albl atb byo时间复杂度： O(N*log N)
辅助空间： O(1)
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