在矩阵中找到单个运动
给定四个整数 x1，y1 和 x2，y2 ，它们代表无限 2D 矩阵中的两个位置，任务是找出是否有可能在一次移动中从(x1，y1)移动到(x2，y2)，或者是向左，向右，向上或向下。请注意，移动将重复进行，直到到达目的地。如果无法达到(x2，y2)输出 -1 。 举例:
输入: x1 = 0，y1 = 0，x2 = 1，y2 = 0 输出:下降 目的地刚好在起点下方。 输入: x1 = 0，y1 = 0，x2 = 1，y2 = 1 输出: -1 单次移动不可能从(0，0)到达(1，1)。
进近:检查坐标是在同一行还是同一列，然后只有可能到达最终目的地。然后根据目的地的方向打印移动。 以下是上述办法的实施情况:
C++
#include <bits/stdc++.h>
using namespace std;
// Function that checks whether it is
// possible to move from
// (x1, y1) to (x2, y2)
void Robot_Grid(int x1, int y1, int x2, int y2)
{
// Both locations are
// in the same row
if (x1 == x2) {
// Destination is
// at the right
if (y1 < y2) {
cout << "Right";
}
// Destination is
// at the left
else {
cout << "Left";
}
}
// Both locations are
// in the same column
else if (y1 == y2) {
// Destination is below
// the current row
if (x1 < x2) {
cout << "Down";
}
// Destination is above
// the current row
else {
cout << "Up";
}
}
// Impossible to get
// to the destination
else {
cout << "-1";
}
}
// Driver code
int main()
{
int x1, x2, y1, y2;
x1 = 0;
y1 = 0;
x2 = 0;
y2 = 1;
Robot_Grid(x1, y1, x2, y2);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the given above approach
public class GFG{
// Function that checks whether it is
// possible to move from
// (x1, y1) to (x2, y2)
static void Robot_Grid(int x1, int y1, int x2, int y2)
{
// Both locations are
// in the same row
if (x1 == x2) {
// Destination is
// at the right
if (y1 < y2) {
System.out.print("Right");
}
// Destination is
// at the left
else {
System.out.print("Left");
}
}
// Both locations are
// in the same column
else if (y1 == y2) {
// Destination is below
// the current row
if (x1 < x2) {
System.out.print("Down");
}
// Destination is above
// the current row
else {
System.out.println("Up");
}
}
// Impossible to get
// to the destination
else {
System.out.print("-1");
}
}
// Driver code
public static void main(String []args)
{
int x1, x2, y1, y2;
x1 = 0;
y1 = 0;
x2 = 0;
y2 = 1;
Robot_Grid(x1, y1, x2, y2);
}
// This code is contributed by Ryuga
}
Python 3
# Function that checks whether it is
# possible to move from
# (x1, y1) to (x2, y2)
def Robot_Grid(x1, y1, x2, y2):
# Both locations are in the same row
if (x1 == x2):
# Destination is at the right
if (y1 < y2):
print("Right")
# Destination is at the left
else:
print("Left")
# Both locations are in the same column
elif (y1 == y2):
# Destination is below the current row
if (x1 < x2):
print("Down")
# Destination is above the current row
else:
print("Up")
# Impossible to get to the destination
else:
print("-1")
# Driver code
if __name__ == '__main__':
x1 = 0
y1 = 0
x2 = 0
y2 = 1
Robot_Grid(x1, y1, x2, y2)
# This code is contributed by
# Sanjit_Prasad
C
// C# implementation of the given above approach
using System;
class GFG{
// Function that checks whether it is
// possible to move from
// (x1, y1) to (x2, y2)
static void Robot_Grid(int x1, int y1,
int x2, int y2)
{
// Both locations are
// in the same row
if (x1 == x2)
{
// Destination is
// at the right
if (y1 < y2)
{
Console.Write("Right");
}
// Destination is
// at the left
else
{
Console.Write("Left");
}
}
// Both locations are
// in the same column
else if (y1 == y2)
{
// Destination is below
// the current row
if (x1 < x2)
{
Console.Write("Down");
}
// Destination is above
// the current row
else
{
Console.WriteLine("Up");
}
}
// Impossible to get
// to the destination
else
{
Console.WriteLine("-1");
}
}
// Driver code
public static void Main()
{
int x1, x2, y1, y2;
x1 = 0;
y1 = 0;
x2 = 0;
y2 = 1;
Robot_Grid(x1, y1, x2, y2);
}
}
// This code is contributed by
// Mukul Singh
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the above approach
// Function that checks whether it is
// possible to move from
// (x1, y1) to (x2, y2)
function Robot_Grid($x1, $y1, $x2, $y2)
{
// Both locations are in the same row
if ($x1 == $x2)
{
// Destination is at the right
if ($y1 < $y2)
{
echo "Right";
}
// Destination is at the left
else
{
echo "Left";
}
}
// Both locations are in the same column
else if ($y1 == $y2)
{
// Destination is below the
// current row
if ($x1 < $x2)
{
echo "Down";
}
// Destination is above the current row
else
{
echo "Up";
}
}
// Impossible to get to the destination
else
{
echo "-1";
}
}
// Driver code
$x1 = 0;
$y1 = 0;
$x2 = 0;
$y2 = 1;
Robot_Grid($x1, $y1, $x2, $y2);
// This code is contributed by ita_c
?>
java 描述语言
<script>
// Javascript implementation of the given above approach
// Function that checks whether it is
// possible to move from
// (x1, y1) to (x2, y2)
function Robot_Grid(x1,y1,x2,y2)
{
// Both locations are
// in the same row
if (x1 == x2) {
// Destination is
// at the right
if (y1 < y2) {
document.write("Right");
}
// Destination is
// at the left
else {
document.write("Left");
}
}
// Both locations are
// in the same column
else if (y1 == y2) {
// Destination is below
// the current row
if (x1 < x2) {
document.write("Down");
}
// Destination is above
// the current row
else {
document.write("Up");
}
}
// Impossible to get
// to the destination
else {
document.write("-1");
}
}
// Driver code
let x1, x2, y1, y2;
x1 = 0;
y1 = 0;
x2 = 0;
y2 = 1;
Robot_Grid(x1, y1, x2, y2);
// This code is contributed by rag2127
</script>
Output:
Right
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