【leetcode】1041. Robot Bounded In Circle

题目如下：

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 degress 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:
Input: "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.
Example 2:
Input: "GG"
Output: false
Explanation:
The robot moves north indefinetely.
Example 3:
Input: "GL"
Output: true
Explanation:
The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ...
Note:

1 <= instructions.length <= 100

instructions[i] is in {'G', 'L', 'R'}

解题思路：看到这个题目，我的感觉就是如果能回到起点，应该是执行instructions一次，两次或者四次。嘿嘿，当然我也不知道怎么证明，反正能AC。

代码如下：

class Solution(object):def process(self,start,instructions):for i in instructions:if i == 'G':if start[2] == 'N':start[1] += 1elif start[2] == 'S':start[1] -= 1elif start[2] == 'E':start[0] += 1elif start[2] == 'W':start[0] -= 1elif i == 'L':if start[2] == 'N':start[2] = 'W'elif start[2] == 'S':start[2] = 'E'elif start[2] == 'E':start[2] = 'N'elif start[2] == 'W':start[2] = 'S'elif i == 'R':if start[2] == 'N':start[2] = 'E'elif start[2] == 'S':start[2] = 'W'elif start[2] == 'E':start[2] = 'S'elif start[2] == 'W':start[2] = 'N'return startdef isRobotBounded(self, instructions):""":type instructions: str:rtype: bool"""start = [0,0,'N']end = self.process(start,instructions)if end[0] == end[1] == 0:return Trueend = self.process(start, instructions)if end[0] == end[1] == 0:return Trueend = self.process(start, instructions)end = self.process(start, instructions)if end[0] == end[1] == 0:return Truereturn False

转载于:https://www.cnblogs.com/seyjs/p/10853771.html

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