通过根据相应的字符串值将位置插入数组来生成序列

给定一个长度为N的字符串S。该字符串仅由字母'F'和'B'组成。任务是生成一个执行某些操作的序列，例如：

• 考虑一个仅由一个 0 组成的整数序列A ，即 A = (0)。
• 现在，对于字符串（1 到 N）的每个索引（i），如果S[i]是“F”，则将i添加到序列A中i-1的最前面（即左侧）
• 否则，如果S[i]是'B'，则将i添加到i-1序列A的紧靠后面（即右侧）。
• 打印结果序列A 。

例子：

Input: N = 5, S = “FBBFB”
Output: 1 2 4 5 3 0
Explanation: Initially, A = {0}.
S[1] is ‘F’ , sequence becomes {1, 0}
S[2] is ‘B’ , sequence becomes {1, 2, 0}
S[3] is ‘B’ , sequence becomes {1, 2, 3, 0}
S[4] is ‘F’ , sequence becomes {1, 2, 4, 3, 0}
S[5] is ‘B’ , sequence becomes {1, 2, 4, 5, 3, 0}

Input : N = 6 , S = “BBBBBB”
Output : 0 1 2 3 4 5 6

方法：解决问题的想法是基于出队的概念。

As at each iteration, it is possible that insertion of i may occur from any end of (i-1), therefore deque can be used as in dequeue insertion is possible from any end. 

按照解决方案的步骤：

• 该方法的观察结果是，在反转所有操作后问题变得更加简单。
• 现在问题修改为，给定的序列仅包含 (N)，并且在从字符串末尾开始的每次迭代之后，
  • 如果S[i]是 'F'，推 i在i-1的右边，
  • 如果S[i]是 'B'，则将i推到 dequeue 中i-1的左侧。
• 从前端开始按顺序返回dequeue的元素。

下面是上述方法的实现。

C++

// C++ program for above approach
 
#include 
using namespace std;
 
// Function to find sequence
// from given string
// according to given rule
void findSequence(int N, string S)
{
 
    // Creating a deque
    deque v;
 
    // Inserting N (size of string) into deque
    v.push_back(N);
 
    // Iterating string from behind and
    // pushing the indices into the deque
    for (int i = N - 1; i >= 0; i--) {
 
        // If letter at current index is 'F',
        // push i to the right of i-1
        if (S[i] == 'F') {
            v.push_back(i);
        }
 
        // If letter at current index is 'B',
        // push i to the left of i-1
        else {
            v.push_front(i);
        }
    }
 
    // Printing resultant sequence
    for (int i = 0; i <= N; i++)
        cout << v[i] << " ";
}
 
// Driver Code
int main()
{
    int N = 5;
    string S = "FBBFB";
 
    // Printing the sequence
    findSequence(N, S);
    return 0;
}

Java

// JAVA program for above approach
import java.util.*;
class GFG
{
 
  // Function to find sequence
  // from given string
  // according to given rule
  public static void findSequence(int N, String S)
  {
 
    // Creating a deque
    Deque v = new ArrayDeque();
 
    // Inserting N (size of string) into deque
    v.addLast(N);
 
    // Iterating string from behind and
    // pushing the indices into the deque
    for (int i = N - 1; i >= 0; i--) {
 
      // If letter at current index is 'F',
      // push i to the right of i-1
      if (S.charAt(i) == 'F') {
        v.addLast(i);
      }
 
      // If letter at current index is 'B',
      // push i to the left of i-1
      else {
        v.addFirst(i);
      }
    }
 
    // Printing resultant sequence
    for (Iterator itr = v.iterator(); itr.hasNext();) {
      System.out.print(itr.next() + " ");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 5;
    String S = "FBBFB";
 
    // Printing the sequence
    findSequence(N, S);
  }
}
 
// This code is contributed by Taranpreet

Python3

# Python program for above approach
from collections import deque
 
# Function to find sequence
# from given string
# according to given rule
def findSequence(N, S):
 
    # Creating a deque
    v = deque()
 
    # Inserting N (size of string) into deque
    v.append(N)
 
    # Iterating string from behind and
    # pushing the indices into the deque
    i = N - 1
    while(i >= 0):
 
        # If letter at current index is 'F',
        # push i to the right of i-1
        if (S[i] == 'F'):
            v.append(i)
 
        # If letter at current index is 'B',
        # push i to the left of i-1
        else:
            v.appendleft(i)
 
        i -= 1
 
    # Printing resultant sequence
    print(*v)
 
# Driver Code
N = 5
S = "FBBFB"
 
# Printing the sequence
findSequence(N, S)
 
# This code is contributed by Samim Hossain Mondal.

C#

// C# program for above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // Function to find sequence
  // from given string
  // according to given rule
  public static void findSequence(int N, String S)
  {
 
    // Creating a deque
    List v = new List();
 
    // Inserting N (size of string) into deque
    v.Add(N);
 
    // Iterating string from behind and
    // pushing the indices into the deque
    for (int i = N - 1; i >= 0; i--) {
 
      // If letter at current index is 'F',
      // push i to the right of i-1
      if (S[i] == 'F') {
        v.Add(i);
      }
 
      // If letter at current index is 'B',
      // push i to the left of i-1
      else {
        v.Insert(0,i);
      }
    }
 
    // Printing resultant sequence
    foreach (int itr in v) {
      Console.Write(itr + " ");
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int N = 5;
    String S = "FBBFB";
 
    // Printing the sequence
    findSequence(N, S);
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

    // JavaScript program for above approach
 
 
    // Function to find sequence
    // from given string
    // according to given rule
    const findSequence = (N, S) => {
 
        // Creating a deque
        let v = [];
 
        // Inserting N (size of string) into deque
        v.push(N);
 
        // Iterating string from behind and
        // pushing the indices into the deque
        for (let i = N - 1; i >= 0; i--) {
 
            // If letter at current index is 'F',
            // push i to the right of i-1
            if (S[i] == 'F') {
                v.push(i);
            }
 
            // If letter at current index is 'B',
            // push i to the left of i-1
            else {
                v.unshift(i);
            }
        }
 
        // Printing resultant sequence
        for (let i = 0; i <= N; i++)
            document.write(`${v[i]} `);
    }
 
    // Driver Code
 
    let N = 5;
    let S = "FBBFB";
 
    // Printing the sequence
    findSequence(N, S);
 
// This code is contributed by rakeshsahni
 

输出

1 2 4 5 3 0 

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