求给定每个球的初始方向的球之间发生的碰撞总数

给定 N 个球在一条线上。每个球的初始方向由仅由“L”和“R”组成的字符串表示，分别代表左右方向。假设碰撞后两个球都改变了方向，碰撞前后速度将保持不变。计算发生碰撞的总数。
例子：

Input: str = "RRLL"
Output: 4

Explanation
After first collision: RLRL
After second collision: LRRL
After third collision: LRLR
After fourth collision: LLRR

Input: str = "RRLR"
Output:  2

Explanation
After first collision: RLRR
After second collision: LRRR

方法：
在每个阶段，我们必须找到子串“RL”的编号，将子串“RL”更改为“LR”，并再次计算结果字符串中子串“RL”的编号，然后重复以下直到没有子串“ RL”可用。我们不会在每个阶段计算子串“RL”的数量，因为它会消耗大量时间。另一种方法是，观察我们在每个阶段看到的结果字符串->“RRLL”->“RLRL”->“LRLR”->“LLRR”。
初始字符串是“RRLL”。让我们有一个方向为 L 的第三个球。现在观察这个 L 正在向字符串的左侧移动。当我们将“RL”交换为“LR”时，我们将把 L 移向完整字符串的左侧。现在类似地，如果我们继续这种交换，这个 L 将面对出现在该字符串左侧的每个 R，因此再次形成“RL”。因此，对于这个 L，“RL”的总数将是出现在该特定 L 左侧的 R 的总数。我们将对每个 L 重复此操作。因此，一般来说，为了计算每一个可能的在每个阶段的子串“RL”，我们将计算每个 L 左侧存在的 R 的总数，因为这些 R 和 L 将在每个阶段形成“RL”，这可以以线性时间复杂度完成。
下面是上述方法的实现：

C++

// C++ implementation to Find total
// no of collisions taking place between
// the balls in which initial direction
// of each ball is given
 
#include 
using namespace std;
 
// Function to count no of collision
int count(string s)
{
    int N, i, cnt = 0, ans = 0;
 
    // length of the string
    N = s.length();
 
    for (i = 0; i < N; i++) {
        if (s[i] == 'R')
            cnt++;
 
        if (s[i] == 'L')
            ans += cnt;
    }
 
    return ans;
}
 
// Driver code
int main()
{
 
    string s = "RRLL";
 
    cout << count(s) << endl;
 
    return 0;
}

Java

// Java implementation to Find total
// no of collisions taking place between
// the balls in which initial direction
// of each ball is given
import java.io.*;
class GFG {
      
// Function to count
// no of collision
static int count(String s)
{
  int N, i, cnt = 0, ans = 0;
 
  // length of the string
  N = s.length();
 
  for (i = 0; i < N; i++)
  {
    if (s.charAt(i) == 'R')
      cnt++;
 
    if (s.charAt(i) == 'L')
      ans += cnt;
  }
  return ans;
}
  
// Driver code
static public void main(String[] args)
{
  String s = "RRLL";
  System.out.println(count(s));
}
}
 
// This code is contributed by Rutvik_56

Python3

# Python3 implementation to find total
# no of collisions taking place between
# the balls in which initial direction
# of each ball is given
 
# Function to count no of collision
def count(s):
 
    cnt, ans = 0, 0
 
    # Length of the string
    N = len(s)
 
    for i in range(N):
        if (s[i] == 'R'):
            cnt += 1
 
        if (s[i] == 'L'):
            ans += cnt
     
    return ans
 
# Driver code
s = "RRLL"
 
print( count(s))
 
# This code is contributed by chitranayal

C#

// C# implementation to Find total
// no of collisions taking place between
// the balls in which initial direction
// of each ball is given
using System;
 
class GFG{
 
// Function to count no of collision
static int count(String s)
{
    int N, i, cnt = 0, ans = 0;
 
    // length of the string
    N = s.Length;
 
    for(i = 0; i < N; i++)
    {
        if (s[i] == 'R')
            cnt++;
 
        if (s[i] == 'L')
            ans += cnt;
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "RRLL";
 
    Console.Write(count(s));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

  // Javascript implementation to Find total
  // no of collisions taking place between
  // the balls in which initial direction
  // of each ball is given
   
  // Function to count no of collision
  function count(s)
  {
      let N, i, cnt = 0, ans = 0;
 
      // length of the string
      N = s.length;
 
      for (i = 0; i < N; i++) {
          if (s[i] == 'R')
              cnt++;
 
          if (s[i] == 'L')
              ans += cnt;
      }
 
      return ans;
  }
   
  let s = "RRLL";
  
  document.write(count(s));
 
// This code is contributed by divyesh072019.

输出：

4

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