从(0，0)开始访问Matrix的每个单元格后的最终方向

给定大小为N x M的2D网格。任务是在给定条件下访问每个单元之后找到最终方向。

• 您只能从N * M网格的左上角开始，并朝右。
• 您可以朝着您的方向一次走一格。
• 如果到达网格的边界，或者已经访问过您要访问的下一个正方形，则向右转。

例子：

Input: N = 3, M = 3 
Output: Right 
Explanation: Below is the final position after traversing the grid 
 

Input: N = 3, M = 1 
Output: Down 
 

方法：对于上面的问题陈述，我们必须注意以下几点：

• 形成的路径将始终是螺旋形路径。因此，我们可以说任何中间单元将成为最终单元，我们需要找到该单元的方向。
• 如果n> m ，则最终方向只能取决于m的值，是Up或Down，因为在覆盖所有其他列时，最后一个未覆盖的列中总会保留一些单元格。
• 如果n <= m ，则取决于n的值，最终方向只能是Left或Right。

因此，根据以上观察结果，四个方向只能有4种情况：

1. 如果n> m且m为偶数，则最终方向将为Up。
2. 如果n> m并且m为奇数，则最终方向将为Down。
3. 如果n <= m且n为偶数，则最终方向将为Left。
4. 如果n <= m并且n为奇数，则最终方向将为Right。

下面是上述方法的实现：

C++

// C++ program to find the direction
// when stopped moving
 
#include 
using namespace std;
 
// Function to find the direction
// when stopped moving
void findDirection(int n, int m)
{
    if (n > m) {
        if (m % 2 == 0)
            printf("Up\n");
        else
            printf("Down\n");
    }
    else {
        if (n % 2 == 0)
            printf("Left\n");
        else
            printf("Right\n");
    }
}
 
// Driver Code
int main()
{
    // Given size of NxM grid
    int n = 3, m = 3;
 
    // Function Call
    findDirection(n, m);
    return 0;
}

Java

// Java program to find the direction
// when stopped moving
class GFG{
 
// Function to find the direction
// when stopped moving
static void findDirection(int n, int m)
{
    if (n > m)
    {
        if (m % 2 == 0)
            System.out.print("Up\n");
        else
            System.out.print("Down\n");
    }
    else
    {
        if (n % 2 == 0)
            System.out.print("Left\n");
        else
            System.out.print("Right\n");
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given size of NxM grid
    int n = 3, m = 3;
 
    // Function Call
    findDirection(n, m);
}
}
 
// This code is contributed by shubham

Python3

# Python3 program to find the direction
# when stopped moving
 
# Function to find the direction
# when stopped moving
def findDirection(n, m):
 
    if (n > m):
        if (m % 2 == 0):
            print("Up\n");
        else:
            print("Down\n");
     
    else:
        if (n % 2 == 0):
            print("Left\n");
        else:
            print("Right\n");
     
# Driver Code
 
# Given size of NxM grid
n = 3; m = 3;
 
# Function Call
findDirection(n, m);
 
# This code is contributed by Code_Mech

C#

// C# program to find the direction
// when stopped moving
using System;
class GFG{
 
// Function to find the direction
// when stopped moving
static void findDirection(int n, int m)
{
    if (n > m)
    {
        if (m % 2 == 0)
            Console.Write("Up\n");
        else
            Console.Write("Down\n");
    }
    else
    {
        if (n % 2 == 0)
            Console.Write("Left\n");
        else
            Console.Write("Right\n");
    }
}
 
// Driver code
public static void Main()
{
     
    // Given size of NxM grid
    int n = 3, m = 3;
 
    // Function Call
    findDirection(n, m);
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript program to find the direction
// when stopped moving
 
// Function to find the direction
// when stopped moving
function findDirection(n, m)
{
    if (n > m) {
        if (m % 2 == 0)
            document.write("Up<br>");
        else
            document.write("Down<br>");
    }
    else {
        if (n % 2 == 0)
            document.write("Left<br>");
        else
            document.write("Right<br>");
    }
}
 
// Driver Code
// Given size of NxM grid
let n = 3, m = 3;
 
// Function Call
findDirection(n, m);
 
// This code is contributed by rishavmahato348.

输出：

Right

时间复杂度： O(1)
辅助空间： O(1)