修改后字符串的最终状态

给定长度为n的字符串S表示彼此相邻的n个盒子。

1. 索引i处的字符R表示第i个框被推向右侧。
2. 另一方面，索引i处的L表示第i个框被推向左侧。
3. 还有一个。代表一个空白空间。

我们从初始配置开始，在每个时间单位上，将一个盒子向右推都会将下一个盒子推向右边，对于将一个盒子推向左边也是如此。任务是在无法再移动时打印所有盒子的最终位置。如果存在子字符串“ RL”，则正从两个方向推动中间框，因此不再移动。

例子：

Input: S = “RR.L” 
Output: RR.L 
The first and second boxes are being toward right. The middle dot is being pushed from both directions, so does not move.

Input: S = “R..R…L.” 
Output: RRRRR.LL. 
At time unit 1 : 
The first box pushes second box. 
Third box pushes fourth box. 
Second last box pushes third last box, configuration becomes 
“RR.RR.LL.” 
At time unit 2 : 
Second box pushes third box, string becomes 
“RRRRR.LL.” 

方法：我们可以计算施加在每个盒子上的净力。我们关心的力是盒子离左R或右L的距离，即距离我们越近，该力越强。

• 从左向右扫描，力在每次迭代中衰减1 ，如果再次遇到R ，则力重置为N。
• 同样，从右到左方向，我们可以从右(接近L )找到力。
• 对于某些盒结果[i]，如果力相等，则结果为。双方都施加相同的力量。否则，无论哪个力强，都暗示着我们的结果。

For string S = “R..R…L.” 
The forces going from left to right are [7, 6, 5, 7, 6, 5, 4, 0, 0]. 
The forces going from right to left are [0, 0, 0, 0, -4, -5, -6, -7, 0]. 
Combining them (taking their arithmetic addition at each index) gives result = [7, 6, 5, 7, 2, 0, -2, -7, 0]. 
Thus, the final answer is RRRRR.LL. 
 

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to return final
// positions of the boxes
string pushBoxes(string S)
{
    int N = S.length();
    vector force(N, 0);
 
    // Populate forces going
    // from left to right
    int f = 0;
    for(int i = 0; i < N; i++)
    {
        if (S[i] == 'R')
        {
            f = N;
        }
        else if (S[i] == 'L')
        {
            f = 0;
        }
        else
        {
            f = max(f - 1, 0);
        }
        force[i] += f;
    }
 
    // Populate forces going from right to left
    f = 0;
    for(int i = N - 1; i >= 0; i--)
    {
        if (S[i] == 'L')
        {
            f = N;
        }
        else if (S[i] == 'R')
        {
            f = 0;
        }
        else
        {
            f = max(f - 1, 0);
        }
        force[i] -= f;
    }
 
    // return final state of boxes
    string ans;
    for(int f : force)
    {
        ans += f == 0 ? '.' : f > 0 ? 'R' : 'L';
    }
    return ans;
}
 
// Driver code
int main()
{
    string S = ".L.R...LR..L..";
     
    // Function call to print answer
    cout << pushBoxes(S);
}
 
// This code is contributed by sanjeev2552

Python3

# Python3 implementation of above approach
 
# Function to return final
# positions of the boxes
def pushBoxes(S):
 
    N = len(S)
    force = [0] * N
 
    # Populate forces going from left to right
    f = 0
    for i in range(0, N):
 
        if S[i] == 'R':
            f = N
        elif S[i] == 'L':
            f = 0
        else:
            f = max(f - 1, 0)
        force[i] += f
 
    # Populate forces going from right to left
    f = 0
    for i in range(N - 1, -1, -1):
 
        if S[i] == 'L':
            f = N
        elif S[i] == 'R':
            f = 0
        else:
            f = max(f - 1, 0)
        force[i] -= f
 
    # return final state of boxes
    return "".join('.' if f == 0 else 'R' if f > 0 else 'L'
                   for f in force)
 
 
# Driver code
S = ".L.R...LR..L.."
 
# Function call to print answer
print(pushBoxes(S))

输出：

LL.RR.LLRRLL..