<div>
On an infinite plane, a robot initially stands at (0, 0) and faces north. The robot can receive one of three instructions:
"G": go straight 1 unit;"L": turn 90 degrees to the left;"R": turn 90 degrees to the right.
The robot performs the instructions given in order, and repeats them forever.
Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.
Example 1:
<pre>Input: instructions = "GGLLGG" Output: true Explanation: The robot moves from (0,0) to (0,2), turns 180 degrees, and then returns to (0,0). When repeating these instructions, the robot remains in the circle of radius 2 centered at the origin.</pre>
Example 2:
<pre>Input: instructions = "GG" Output: false Explanation: The robot moves north indefinitely.</pre>
Example 3:
<pre>Input: instructions = "GL" Output: true Explanation: The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ...</pre>
Constraints:
1 <= instructions.length <= 100instructions[i]is'G','L'or,'R'.
</div>
题目大意：
循环按模式走是否回到原点
解题思路：
数学题，很难证明。定理是，只要按照给定模式走完，若回到原点或最后方向不是向北，都能回到原点
解题步骤：
N/A
注意事项：
Python代码：
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22def isRobotBounded(self, instructions: str) -> bool:
DIRECTION_CONVERT_LEFT = {
(0, 1): (-1, 0),
(-1, 0): (0, -1),
(0, -1): (1, 0),
(1, 0): (0, 1),
}
DIRECTION_CONVERT_RIGHT = {
(0, 1): (1, 0),
(1, 0): (0, -1),
(0, -1): (-1, 0),
(-1, 0): (0, 1),
}
path, direction, position = instructions, (0, 1), (0, 0)
for char in path:
if char == 'L':
direction = DIRECTION_CONVERT_LEFT[direction]
elif char == 'R':
direction = DIRECTION_CONVERT_RIGHT[direction]
else:
position = (position[0] + direction[0], position[1] + direction[1])
return True if position == (0, 0) or direction != (0, 1) else False
算法分析：
时间复杂度为O(n)，空间复杂度O(1)
