【leetcode】1041. Robot Bounded In Circle

最新推荐文章于 2025-08-15 16:44:13 发布

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最新推荐文章于 2025-08-15 16:44:13 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/seyjs/p/10853771.html

本文介绍了一个判断机器人在无限平面上是否能被限定在一个圈内的算法。通过解析指令集G前进,L左转90度,R右转90度,分析其循环特性。若机器人最终返回原点或改变方向,则存在限制圈。

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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] += 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