657. Robot Return to Origin
Easy
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'
.
Runtime: 4 ms, faster than 84.29%
Memory Usage: 3.2 MB, less than 85.71%
func judgeCircle(moves string) bool {
x := 0
y := 0
for _, char := range moves {
if char == 'U' {
y++
} else if char == 'D' {
y--
} else if char == 'R' {
x++
} else if char == 'L' {
x--
}
}
return x==0 && y==0
}
Last modified 6mo ago