根据给定的索引重新排列字符串

给定一个字符串S和一个数组index [] ，任务是通过将每个字符S [i]放置到位置index [i]来重新排列字符串S。
例子

Input: S = “geeksforgeeks”, index[] = {5, 6, 7, 0, 1, 2, 8, 9, 10, 3, 4, 11, 12} 
Output: ksfeegeeorgks
Input: S = “math”, index[] = {0, 1, 2, 3} 
Output: math 
 

方法：
要解决此问题，请按照以下步骤操作：

• 字符串S转换为字符的列表，因为字符串在本质不变。
• 复制列表。根据index [i]的值重新排列此列表中的字符。
• 将列表转换为字符串并打印最终字符串。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to convert the strings
// to proper
void Convertstrings(string s, int index[],
                              int n)
{
    char a[s.length()];
    char b[s.length()];
     
    // Convert string to array
    for(int ii = 0; ii < s.length(); ii++)
    {
        a[ii] = s[ii];
        b[ii] = s[ii];
    }
     
    int i = 0, j = 0;
     
    // Move characters to specified indices
    while(j < s.length() && i < n)
    {
        int k = index[i];
        int temp = a[j];
        b[k] = temp;
         
        j += 1;
        i += 1;
    }
     
    string tmp = "";
     
    // Convert the list to string
    for(i = 0; i < s.length(); i++)
    {
        tmp += b[i];
    }
 
    // Print the answer
    cout << tmp << endl;
}
 
// Driver Code
int main()
{
    string s = "geeksforgeeks";
    int index[] = { 5, 6, 7, 0, 1, 2, 8,
                    9, 10, 3, 4, 11, 12};
     
    int n = sizeof(index) / sizeof(index[0]);
     
    Convertstrings(s, index, n);
    return 0;
}
 
// This code is contributed by rutvik_56

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to convert the Strings
// to proper
static void ConvertStrings(char []s,
                           int index[],
                           int n)
{
    char []a = new char[s.length];
    char []b = new char[s.length];
     
    // Convert String to array
    for(int ii = 0; ii < s.length; ii++)
    {
        a[ii] = s[ii];
        b[ii] = s[ii];
    }
     
    int i = 0, j = 0;
     
    // Move characters to specified indices
    while(j < s.length && i < n)
    {
        int k = index[i];
        int temp = a[j];
        b[k] = (char) temp;
         
        j += 1;
        i += 1;
    }
     
    String tmp = "";
     
    // Convert the list to String
    for(i = 0; i < s.length; i++)
    {
        tmp += b[i];
    }
 
    // Print the answer
    System.out.print(tmp +"\n");
}
 
// Driver Code
public static void main(String[] args)
{
    String s = "geeksforgeeks";
    int index[] = { 5, 6, 7, 0, 1, 2, 8,
                    9, 10, 3, 4, 11, 12};
     
    int n = index.length;
     
    ConvertStrings(s.toCharArray(), index, n);
}
}
 
// This code is contributed by Rohit_ranjan

Python3

# Python3 Program to implement
# the above approach
 
# Function to convert the strings
# to proper
def Convertstrings(s, index):
    a = []
    j = 0
    i = 0
 
    # Convert string to list
    for ii in str(s):
        a.append(ii)
 
    # Copy the list to another list
    b = a[:]
 
    # Move characters to specified indices
    while j < len(a) and i < len(index):
        k = index[i]
        temp = a[j]
        b[k] = temp
        j += 1
        i += 1
    s = ''
 
    # Convert the list to string
    for i in range(len(b)):
        s += b[i]
 
    # Print the answer
    print(s)
 
 
# Driver Code
s = "geeksforgeeks"
index = [5, 6, 7, 0, 1, 2, 8, 9, 10, 3, 4, 11, 12]
Convertstrings(s, index)

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to convert the Strings
// to proper
static void ConvertStrings(char []s,
                           int []index,
                           int n)
{
    char []a = new char[s.Length];
    char []b = new char[s.Length];
     
    // Convert String to array
    for(int ii = 0; ii < s.Length; ii++)
    {
        a[ii] = s[ii];
        b[ii] = s[ii];
    }
     
    int i = 0, j = 0;
     
    // Move characters to specified indices
    while(j < s.Length && i < n)
    {
        int k = index[i];
        int temp = a[j];
        b[k] = (char) temp;
         
        j += 1;
        i += 1;
    }
     
    String tmp = "";
     
    // Convert the list to String
    for(i = 0; i < s.Length; i++)
    {
        tmp += b[i];
    }
 
    // Print the answer
    Console.Write(tmp +"\n");
}
 
// Driver Code
public static void Main(String[] args)
{
    String s = "geeksforgeeks";
    int []index = { 5, 6, 7, 0, 1, 2, 8,
                    9, 10, 3, 4, 11, 12};
     
    int n = index.Length;
     
    ConvertStrings(s.ToCharArray(), index, n);
}
}
 
// This code is contributed by Rajput-Ji

输出：

ksfeegeeorgks

时间复杂度： O(N)
辅助空间： O(N)