[Swift]LeetCode1041. 困于环中的机器人 | Robot Bounded In Circle

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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'}

---

在无限的平面上，机器人最初位于 (0, 0) 处，面朝北方。机器人可以接受下列三条指令之一：

• "G"：直走 1 个单位
• "L"：左转 90 度
• "R"：右转 90 度

机器人按顺序执行指令 instructions，并一直重复它们。

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。

示例 1：

输入："GGLLGG"
输出：true
解释：
机器人从 (0,0) 移动到 (0,2)，转 180 度，然后回到 (0,0)。
重复这些指令，机器人将保持在以原点为中心，2 为半径的环中进行移动。

示例 2：

输入："GG"
输出：false
解释：
机器人无限向北移动。

示例 3：

输入："GL"
输出：true
解释：
机器人按 (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ... 进行移动。

提示：

1. 1 <= instructions.length <= 100
2. instructions[i] 在 {'G', 'L', 'R'} 中

---

0ms

class Solution {
    enum dir {
        case px, py, nx, ny
        
        func turnL() -> dir
        {
            switch self {
            case .px: return .py
            case .py: return .nx
            case .nx: return .ny
            case .ny: return .px
            }
        }
        
        func trunR() -> dir
        {
            switch self {
            case .px: return .ny
            case .ny: return .nx
            case .nx: return .py
            case .py: return .px
            }
        }
        
        func move(point: inout [Int])
        {
            switch self {
            case .px: point[0] += 1; return
            case .py: point[1] += 1; return
            case .nx: point[0] -= 1; return
            case .ny: point[1] -= 1; return
            }
        }
    }
    
    func isRobotBounded(_ instructions: String) -> Bool
    {
        if self.help(instructions) { return true }
        else
        {
            var newInput = instructions
            for _ in 1...3
            {
                newInput.append(instructions)
            }
            return self.help(newInput)
        }
    }
    
    private func help(_ instructions: String) -> Bool
    {
        var d: dir = .px
        // index0: x, index1: y
        var p = [0,0]
        
        for c in instructions
        {
            if c == "G" { d.move(point: &p) }
            else if c == "L" { d = d.turnL() }
            else { d = d.trunR() }
        }
        
        return p[0] == 0 && p[1] == 0
    }
}

---

Runtime: 4 ms

Memory Usage: 20.5 MB

class Solution {
    func isRobotBounded(_ instructions: String) -> Bool {
        let dx:[Int] = [0,-1,0,1]
        let dy:[Int] = [1,0,-1,0]
        let arr:[Character] = Array(instructions)
        var dir:Int = 0
        var x:Int = 0
        var y:Int = 0
        for i in 0..<arr.count
        {
            switch arr[i]
            {
                case "L":
                dir += 1
                case "R":
                dir += 3
                default:
                x += dx[dir]
                y += dy[dir]
            }
            dir %= 4
        }
        return !((x != 0 || y != 0) && dir == 0)
    }
}

---

Runtime: 8 ms

Memory Usage: 20.6 MB

class Solution {
    func isRobotBounded(_ instructions: String) -> Bool {
        let dx:[Int] = [0,-1,0,1]
        let dy:[Int] = [1,0,-1,0]
        let arr:[Character] = Array(instructions)
        var dir:Int = 0
        var x:Int = 0
        var y:Int = 0
        for i in 0..<arr.count
        {
            switch arr[i]
            {
                case "L":
                dir += 1
                case "R":
                dir += 3
                default:
                x += dx[dir]
                y += dy[dir]
            }
            dir %= 4
        }
        if dir != 0 {return true}
        if x == 0 && y == 0 {return true}
        return false
    }
}

---

4ms

class Solution {
    let directions = [(0, 1), (-1, 0), (0, -1), (1, 0)]
    func isRobotBounded(_ instructions: String) -> Bool {

        var index = 0
        var location = (0, 0)
        // for _ in 1...4 {
            location = getOnePass(instructions, location, &index)
        // }
        return location.0 == 0 && location.1 == 0 || index > 0
    }

    fileprivate func getOnePass(_ instructions: String, _ start: (Int, Int), _ index:inout Int) -> (Int, Int) {
        
        var curr = start
        for item in instructions {
            if item == "G" {
                curr = (curr.0 + directions[index].0, curr.1 + directions[index].1)
            } else if item == "L" {
                index = (index + 3) % 4
            } else if item == "R" {
                index = (index + 1) % 4
            }
        }
        return curr
    }
}