描述
There is a robot starting at 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.
The move sequence is represented by a string, and the character moves[i] represents its ith move. Valid moves are R (right), L (left), U (up), and D (down). If the robot returns to the origin after it finishes all of its moves, return true. Otherwise, return false.
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.
Example 3:
Input: moves = "RRDD"
Output: false
Example 4:
Input: moves = "LDRRLRUULR"
Output: false
Note:
1 <= moves.length <= 2 * 104
moves only contains the characters 'U', 'D', 'L' and 'R'.
解析
根据题意，其实只需要判断机器人往上和往下的步子一样，往左和往右的步子一样，机器人就扔在原地。
解答
class Solution(object):
def judgeCircle(self, moves):
"""
:type moves: str
:rtype: bool
"""
if moves=="":
return True
return moves.count('R')==moves.count('L') and moves.count('U')==moves.count('D')
运行结果
Runtime: 16 ms, faster than 99.15% of Python online submissions for Robot Return to Origin.
Memory Usage: 13.7 MB, less than 44.59% of Python online submissions for Robot Return to Origin.
原题链接：leetcode.com/problems/ro…
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