657. Robot Return to Origin(Solution || Leetcode easy || Java)
There is a robot starting at the position (0, 0)
, the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0)
after it completes its moves.
You are given a string moves
that represents the move sequence of the robot where moves[i]
represents its ith
move. Valid moves are 'R'
(right), 'L'
(left), 'U'
(up), and 'D'
(down).
Return true
if the robot returns to the origin after it finishes all of its moves, or false
otherwise.
Note: The way that the robot is “facing” is irrelevant. 'R'
will always make the robot move to the right once, 'L'
will always make it move left, etc. Also, assume that the magnitude of the robot's movement is the same for each move.
Example 1:
Input: moves = "UD"
Output: true
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.
Example 2:
Input: moves = "LL"
Output: false
Explanation: The robot moves left twice. It ends up two "moves" to the left of the origin. We return false because it is not at the origin at the end of its moves.
Constraints:
1 <= moves.length <= 2 * 104
moves
only contains the characters'U'
,'D'
,'L'
and'R'
.
SOLUTION(1):
class Solution {
public boolean judgeCircle(String moves) {
//Initialise variable to count up and down movement
int vertical = 0;
//Initialise variable to count right and left movement
int horizontal = 0;
for(int i = 0; i < moves.length(); i++){
if(moves.charAt(i) == 'U'){
vertical++;
}
else if(moves.charAt(i) == 'D'){
vertical--;
}
else if(moves.charAt(i) == 'R'){
horizontal++;
}
else{
horizontal--;
}
}
return vertical == 0 && horizontal == 0;
}
}
SOLUTION(2):
class Solution {
public boolean judgeCircle(String moves) {
int vertical = 0;
int horizontal = 0;
for(char i : moves.toCharArray()){
switch (i){
case 'U' : vertical++; break;
case 'D' : vertical--; break;
case 'R' : horizontal++; break;
case 'L' : horizontal--; break;
}
}
return vertical == 0 && horizontal == 0;
}
}
Thank you for reading. If you have any queries then, please let me know in the comment section. I will surely be responsive toward it.