根据给定的不等式排列数组中的数字

给定N个不同整数的列表和N-1个不等号的列表，任务是在不等号之间插入整数，以使形成的最终不等式始终成立。
注意：不等号的顺序不得更改。

例子：

Input: Integers: [ 2, 5, 1, 0 ], Signs: [ <, >, < ]
Output: 0 < 5 > 1 < 2 
Explanation:
The inequality formed is consistent and valid.

Input: Integers: [ 8, 34, 25, 1, -5, 10], Signs: [ >,  >, <, <, > ]
Output: 34 > 25 > -5 < 1 < 10 > 8
Explanation:
The inequality formed is consistent and valid.

方法：不等式符号列表可以包含任何顺序的符号。因此，为了获得一致的不等式，请将数组中每个<符号前的最小整数和每个>符号前的最大整数放置在数组中。基于此想法，请执行以下步骤：

1. 按升序对整数列表进行排序。
2. 保持两个变量low和high指向整数列表的第一个和最后一个索引。
3. 遍历不等式符号列表。如果当前符号较小，则在<之前加上low所指向的整数，并递增low以指向下一个索引。如果当前符号较大，则在>之前添加由high指向的整数，并递减high指向先前的索引。
4. 最后，将剩余的元素添加到最后一个位置。

下面是上述方法的实现：

Java

// Java program for the above approach
import java.util.Arrays;
 
public class PlacingNumbers {
 
    // Function to place the integers
    // in between the inequality signs
    static String
    formAnInequality(int[] integers,
                     char[] inequalities)
    {
 
        // Sort the integers array and
        // set the index of smallest
        // and largest element
        Arrays.sort(integers);
 
        int lowerIndex = 0;
        int higherIndex = integers.length - 1;
 
        StringBuilder sb = new StringBuilder();
 
        // Iterate over the inequalities
        for (char ch : inequalities) {
 
            // Append the necessary
            // integers per symbol
            if (ch == '<') {
                sb.append(" "
                          + integers[lowerIndex++]
                          + " "
                          + ch);
            }
            else {
                sb.append(" "
                          + integers[higherIndex--]
                          + " "
                          + ch);
            }
        }
 
        // Add the final integer
        sb.append(" " + integers[lowerIndex]);
 
        // Return the answer
        return sb.toString();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given List of Integers
        int[] integers = { 2, 5, 1, 0 };
 
        // Given list of inequalities
        char[] inequalities = { '<', '>', '<' };
 
        // Function Call
        String output
            = formAnInequality(integers,
                               inequalities);
 
        // Print the output
        System.out.println(output);
    }
}

Python3

# Python3 program for
# the above approach
 
# Function to place the integers
# in between the inequality signs
def formAnInequality(integers,
                     inequalities):
 
    # Sort the integers array and
    # set the index of smallest
    # and largest element
    integers.sort()
 
    lowerIndex = 0
    higherIndex = len(integers) - 1
    sb = ""
 
    # Iterate over the inequalities
    for ch in  inequalities:
 
        # Append the necessary
        # integers per symbol
        if (ch == '<'):
            sb += (" " + chr(integers[lowerIndex]) +
                   " " + ch)
            lowerIndex += 1
        else:
            sb += (" " + chr(integers[higherIndex]) +
                   " " + ch)
            higherIndex -= 1
        
    # Add the final integer
    sb += (" " + chr(integers[lowerIndex]))
 
    # Return the answer
    return sb
 
# Driver Code
if __name__ ==  "__main__":
     
    # Given List of Integers
    integers = [2, 5, 1, 0]
 
    # Given list of inequalities
    inequalities = ['<', '>', '<']
 
    # Function Call
    output = formAnInequality(integers,
                              inequalities)
 
    # Print the output
    print(output)
 
# This code is contributed by Chitranayal

C#

// C# program for the above approach
using System;
using System.Text;
 
class GFG{
 
// Function to place the integers
// in between the inequality signs
static String
formAnInequality(int[] integers,
                char[] inequalities)
{
     
    // Sort the integers array and
    // set the index of smallest
    // and largest element
    Array.Sort(integers);
 
    int lowerIndex = 0;
    int higherIndex = integers.Length - 1;
 
    StringBuilder sb = new StringBuilder();
 
    // Iterate over the inequalities
    foreach(char ch in inequalities)
    {
         
        // Append the necessary
        // integers per symbol
        if (ch == '<')
        {
            sb.Append(" " + integers[lowerIndex++] +
                      " " + ch);
        }
        else
        {
            sb.Append(" " + integers[higherIndex--] +
                      " " + ch);
        }
    }
 
    // Add the readonly integer
    sb.Append(" " + integers[lowerIndex]);
 
    // Return the answer
    return sb.ToString();
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given List of ints
    int[] integers = { 2, 5, 1, 0 };
 
    // Given list of inequalities
    char[] inequalities = { '<', '>', '<' };
 
    // Function call
    String output = formAnInequality(integers,
                                      inequalities);
 
    // Print the output
    Console.WriteLine(output);
}
}
 
// This code is contributed by 29AjayKumar

输出：

0 < 5 > 1 < 2

时间复杂度： O(N * log N)
辅助空间： O(1)