最终细胞在矩阵中的位置

给定命令列表U(Up)，D(Down)，L(Left)和R(Right)以及矩阵中初始单元格位置(x，y)的列表。按照给定的命令查找对象在矩阵中的最终单元格位置。假定最终所需的单元格位置存在于矩阵中。

例子：

Input : command[] = "DDLRULL"
        x = 3, y = 4
Output : (1, 5)

Input : command[] = "LLRUUUDRRDDDULRLLUDUUR"
        x = 6, y = 5
Output : (6, 3)

资料来源：Flipkart访谈(校园内的SDE-1)。

方法：以下是步骤：

1. 分别计算杯子(U)，上(D)，下(D)，左(左)和右(右)运动的杯，下，裂和裂。
2. 计算final_x = x +(深–裂)和final_y = y +(深–杯)。

最终单元格位置是(final_x，final_y)

C++

// C++ implementation to find the final cell position
// in the given matrix
#include 
  
using namespace std;
  
// function to find the final cell position
// in the given matrix
void finalPos(char command[], int n,
              int x, int y)
{
    // to count up, down, left and cright
    // movements
    int cup, cdown, cleft, cright;
  
    // to store the final coordinate position
    int final_x, final_y;
  
    cup = cdown = cleft = cright = 0;
  
    // traverse the command array
    for (int i = 0; i < n; i++) {
        if (command[i] == 'U')
            cup++;
        else if (command[i] == 'D')
            cdown++;
        else if (command[i] == 'L')
            cleft++;
        else if (command[i] == 'R')
            cright++;
    }
  
    // calculate final values
    final_x = x + (cright - cleft);
    final_y = y + (cdown - cup);
  
    cout << "Final Position: "
         << "("
         << final_x << ", " << final_y << ")";
}
  
// Driver program to test above
int main()
{
    char command[] = "DDLRULL";
    int n = (sizeof(command) / sizeof(char)) - 1;
    int x = 3, y = 4;
    finalPos(command, n, x, y);
    return 0;
}

Java

// Java implementation to find 
// the final cell position
// in the given matrix
import java.util.*;
import java.lang.*;
import java.io.*;
  
class GFG
{
// function to find the 
// final cell position
// in the given matrix
static void finalPos(String command, int n,
                              int x, int y)
{
    // to count up, down, left 
    // and cright movements
    int cup, cdown, cleft, cright;
  
    // to store the final
    // coordinate position
    int final_x, final_y;
  
    cup = cdown = cleft = cright = 0;
  
    // traverse the command array
    for (int i = 0; i < n; i++) 
    {
        if (command.charAt(i) == 'U')
            cup++;
        else if (command.charAt(i) == 'D')
            cdown++;
        else if (command.charAt(i)== 'L')
            cleft++;
        else if (command.charAt(i) == 'R')
            cright++;
    }
  
    // calculate final values
    final_x = x + (cright - cleft);
    final_y = y + (cdown - cup);
  
    System.out.println("Final Position: " + 
                       "(" + final_x + ", " + 
                              final_y + ")");
}
  
// Driver Code
public static void main(String []args)
{
    String command = "DDLRULL";
    int n = command.length();
    int x = 3, y = 4;
    finalPos(command, n, x, y);
}
}
  
// This code is contributed
// by Subhadeep

C#

// C# implementation to find 
// the final cell position
// in the given matrix
class GFG
{
// function to find the 
// final cell position
// in the given matrix
static void finalPos(string command, int n,
                              int x, int y)
{
    // to count up, down, left 
    // and cright movements
    int cup, cdown, cleft, cright;
  
    // to store the final
    // coordinate position
    int final_x, final_y;
  
    cup = cdown = cleft = cright = 0;
  
    // traverse the command array
    for (int i = 0; i < n; i++) 
    {
        if (command[i] == 'U')
            cup++;
        else if (command[i] == 'D')
            cdown++;
        else if (command[i]== 'L')
            cleft++;
        else if (command[i] == 'R')
            cright++;
    }
  
    // calculate final values
    final_x = x + (cright - cleft);
    final_y = y + (cdown - cup);
  
    System.Console.WriteLine("Final Position: " + 
                             "(" + final_x + ", " + 
                                    final_y + ")");
}
  
// Driver Code
public static void Main()
{
    string command = "DDLRULL";
    int n = command.Length;
    int x = 3, y = 4;
    finalPos(command, n, x, y);
}
}
  
// This code is contributed
// by mits

---

输出：

Final Position: (1, 5)

时间复杂度：O(n)，其中$n$是命令数。