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LeetCode/1041-1050/1041. 困于环中的机器人(中等).md
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| ### 题目描述 | ||
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| 这是 LeetCode 上的 **[1041. 困于环中的机器人](https://leetcode.cn/problems/robot-bounded-in-circle/solution/gong-shui-san-xie-chang-gui-mo-ni-ti-by-hgdtp/)** ,难度为 **中等**。 | ||
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| Tag : 「模拟」、「脑筋急转弯」 | ||
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| 在无限的平面上,机器人最初位于 $(0, 0)$ 处,面朝北方。注意: | ||
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| * 北方向 是 `y` 轴的正方向。 | ||
| * 南方向 是 `y` 轴的负方向。 | ||
| * 东方向 是 `x` 轴的正方向。 | ||
| * 西方向 是 `x` 轴的负方向。 | ||
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| 机器人可以接受下列三条指令之一: | ||
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| * `"G"`:直走 1ドル$ 个单位 | ||
| * `"L"`:左转 90ドル$ 度 | ||
| * `"R"`:右转 90ドル$ 度 | ||
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| 机器人按顺序执行指令 `instructions`,并一直重复它们。 | ||
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| 只有在平面中存在环使得机器人永远无法离开时,返回 `true`。否则,返回 `false`。 | ||
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| 示例 1: | ||
| ``` | ||
| 输入:instructions = "GGLLGG" | ||
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| 输出:true | ||
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| 解释:机器人最初在(0,0)处,面向北方。 | ||
| "G":移动一步。位置:(0,1)方向:北。 | ||
| "G":移动一步。位置:(0,2).方向:北。 | ||
| "L":逆时针旋转90度。位置:(0,2).方向:西。 | ||
| "L":逆时针旋转90度。位置:(0,2)方向:南。 | ||
| "G":移动一步。位置:(0,1)方向:南。 | ||
| "G":移动一步。位置:(0,0)方向:南。 | ||
| 重复指令,机器人进入循环:(0,0)——>(0,1)——>(0,2)——>(0,1)——>(0,0)。 | ||
| 在此基础上,我们返回true。 | ||
| ``` | ||
| 示例 2: | ||
| ``` | ||
| 输入:instructions = "GG" | ||
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| 输出:false | ||
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| 解释:机器人最初在(0,0)处,面向北方。 | ||
| "G":移动一步。位置:(0,1)方向:北。 | ||
| "G":移动一步。位置:(0,2).方向:北。 | ||
| 重复这些指示,继续朝北前进,不会进入循环。 | ||
| 在此基础上,返回false。 | ||
| ``` | ||
| 示例 3: | ||
| ``` | ||
| 输入:instructions = "GL" | ||
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| 输出:true | ||
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| 解释:机器人最初在(0,0)处,面向北方。 | ||
| "G":移动一步。位置:(0,1)方向:北。 | ||
| "L":逆时针旋转90度。位置:(0,1).方向:西。 | ||
| "G":移动一步。位置:(- 1,1)方向:西。 | ||
| "L":逆时针旋转90度。位置:(- 1,1)方向:南。 | ||
| "G":移动一步。位置:(- 1,0)方向:南。 | ||
| "L":逆时针旋转90度。位置:(- 1,0)方向:东方。 | ||
| "G":移动一步。位置:(0,0)方向:东方。 | ||
| "L":逆时针旋转90度。位置:(0,0)方向:北。 | ||
| 重复指令,机器人进入循环:(0,0)——>(0,1)——>(- 1,1)——>(- 1,0)——>(0,0)。 | ||
| 在此基础上,我们返回true。 | ||
| ``` | ||
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| 提示: | ||
| * 1ドル <= instructions.length <= 100$ | ||
| * `instructions[i]` 仅包含 `'G'`, `'L'`, `'R'` | ||
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| --- | ||
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| ### 模拟 | ||
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| 为了方便,将 `instructions` 记为 `s`,"北西南东"四个方向分别记为"上左下右"四个逆时针方向。 | ||
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| 起始位置在 $(0,0),ドル方向为上,我们可以将「位置 + 方向」统称为「状态」。 | ||
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| 所谓"循环",则是指执行若干次的 `s` 后,会回到相同的状态。 | ||
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| 我们可以按 `s` 执行一遍,假设执行完所在位置为 $(x, y),ドル所在位置为 $k,ドル先根据 **位置** 分情况讨论: | ||
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| 1. $(x, y)$ 为 $(0, 0),ドル此时无论执行一遍后的方向为何值,必然能在若干次执行后回到起始状态。 | ||
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| 即只需要确保 `(n * k) % 4` 为 0ドル$ 即可,机器人会陷入循环; | ||
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| 2. $(x, y)$ 不为 $(0, 0),ドル再根据 **方向** 进一步分情况讨论: | ||
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| * 方向为上:每执行一遍 `s` 指令,位置变化为 $(x, y),ドル方向不变。 | ||
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| 那么执行 $n$ 遍后位置为 $(n \times x, n \times y),ドル其中 $n$ 为正整数,并且 $x$ 和 $y$ 不同时为 0ドル,ドル因此随着执行次数增加,位置会离 $(0, 0)$ 越来越远,机器人不会陷入循环; | ||
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| * 方向为下:每执行一遍 `s` 指令,位置变化为 $(x, y),ドル方向翻转。 | ||
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| 如果再执行一遍,由于再次执行时的方向与起始方向相反,因此位置变化为 $(-x, -y),ドル同时方向再次翻转(与起始方向一致)。即执行偶数次后,会回到起始状态,机器人会陷入循环; | ||
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| * 方向为左:每执行一遍 `s` 指令,位置变化为 $(x, y),ドル方向为逆时针 90ドル$ 度。 | ||
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| 如果执行第二次,位置变化为 $(-y, x),ドル方向再逆时针 90ドル$ 度(与起始方向相反);执行第三次,位置变化为 $(-x, -y),ドル方向再逆时针 90ドル$ 度(与起始方向呈顺时针 90ドル$ 度),该次变化会和首次执行相互抵消;执行第四次,位置变化为 $(y, -x),ドル方向再逆时针 90ドル$ 度(与起始方向相同),该次变化与第二次执行相互抵消。总的位置变化为 $(0, 0),ドル同时方向与起始方向一致,机器人会陷入循环; | ||
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| * 方向为右:与「方向为左」同理,机器人会陷入循环。 | ||
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| 综上,只要执行一遍 `s` 后所在位置为 $(0, 0)$ 或方向不为上,均可确保循环发生。 | ||
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| Java 代码: | ||
| ```Java | ||
| class Solution { | ||
| public boolean isRobotBounded(String s) { | ||
| int x = 0, y = 0, d = 0; | ||
| int[][] dirs = new int[][]{{0,1}, {-1,0}, {0,-1}, {1,0}}; | ||
| for (char c : s.toCharArray()) { | ||
| if (c == 'G') { | ||
| x += dirs[d][0]; y += dirs[d][1]; | ||
| } else if (c == 'L') { | ||
| d = (d + 1) % 4; | ||
| } else { | ||
| d = ((d - 1) % 4 + 4) % 4; | ||
| } | ||
| } | ||
| return (x == 0 && y == 0) || d != 0; | ||
| } | ||
| } | ||
| ``` | ||
| C++ 代码: | ||
| ```C++ | ||
| class Solution { | ||
| public: | ||
| bool isRobotBounded(string s) { | ||
| int x = 0, y = 0, d = 0; | ||
| int dirs[4][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}}; | ||
| for (char c : s) { | ||
| if (c == 'G') { | ||
| x += dirs[d][0]; y += dirs[d][1]; | ||
| } else if (c == 'L') { | ||
| d = (d + 1) % 4; | ||
| } else { | ||
| d = ((d - 1) % 4 + 4) % 4; | ||
| } | ||
| } | ||
| return (x == 0 && y == 0) || d != 0; | ||
| } | ||
| }; | ||
| ``` | ||
| Python 代码: | ||
| ```Python | ||
| class Solution: | ||
| def isRobotBounded(self, s: str) -> bool: | ||
| x, y, d = 0, 0, 0 | ||
| dirs = [[0, 1], [-1, 0], [0, -1], [1, 0]] | ||
| for c in s: | ||
| if c == 'G': | ||
| x += dirs[d][0] | ||
| y += dirs[d][1] | ||
| elif c == 'L': | ||
| d = (d + 1) % 4 | ||
| else: | ||
| d = ((d - 1) % 4 + 4) % 4 | ||
| return (x == 0 and y == 0) or d != 0 | ||
| ``` | ||
| Go 代码: | ||
| ```Go | ||
| func isRobotBounded(s string) bool { | ||
| x, y, d := 0, 0, 0 | ||
| dirs := [][]int{{0, 1}, {-1, 0}, {0, -1}, {1, 0}} | ||
| for _, c := range s { | ||
| if c == 'G' { | ||
| x += dirs[d][0] | ||
| y += dirs[d][1] | ||
| } else if c == 'L' { | ||
| d = (d + 1) % 4 | ||
| } else { | ||
| d = ((d - 1) % 4 + 4) % 4 | ||
| } | ||
| } | ||
| return (x == 0 && y == 0) || d != 0 | ||
| } | ||
| ``` | ||
| TypeScript 代码: | ||
| ```TypeScript | ||
| function isRobotBounded(s: string): boolean { | ||
| let x = 0, y = 0, d = 0; | ||
| const dirs: number[][] = [[0, 1], [-1, 0], [0, -1], [1, 0]]; | ||
| for (const c of s) { | ||
| if (c === 'G') { | ||
| x += dirs[d][0]; | ||
| y += dirs[d][1]; | ||
| } else if (c === 'L') { | ||
| d = (d + 1) % 4; | ||
| } else { | ||
| d = ((d - 1) % 4 + 4) % 4; | ||
| } | ||
| } | ||
| return (x === 0 && y === 0) || d !== 0; | ||
| }; | ||
| ``` | ||
| * 时间复杂度:$O(n)$ | ||
| * 空间复杂度:$O(1)$ | ||
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| --- | ||
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| ### 最后 | ||
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| 这是我们「刷穿 LeetCode」系列文章的第 `No.1041` 篇,系列开始于 2021年01月01日,截止于起始日 LeetCode 上共有 1916 道题目,部分是有锁题,我们将先把所有不带锁的题目刷完。 | ||
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| 在这个系列文章里面,除了讲解解题思路以外,还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。 | ||
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| 为了方便各位同学能够电脑上进行调试和提交代码,我建立了相关的仓库:https://github.com/SharingSource/LogicStack-LeetCode 。 | ||
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| 在仓库地址里,你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。 | ||
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