【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. 1 <= instructions.length <= 100
2. 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] += 1
                elif start[2] == 'S':start[1] -= 1
                elif start[2] == 'E':start[0] += 1
                elif start[2] == 'W':start[0] -= 1
            elif 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 start

    def isRobotBounded(self, instructions):
        """
        :type instructions: str
        :rtype: bool
        """
        start = [0,0,'N']
        end = self.process(start,instructions)
        if end[0] == end[1] == 0:
            return True
        end = self.process(start, instructions)
        if end[0] == end[1] == 0:
            return True
        end = self.process(start, instructions)
        end = self.process(start, instructions)
        if end[0] == end[1] == 0:
            return True
        return False

标签：elif,end,self,1041,robot,start,Bounded,leetcode,instructions
来源： https://www.cnblogs.com/seyjs/p/10853771.html