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Formatted question description: https://leetcode.ca/all/1041.html
1041. Robot Bounded In Circle (Medium)
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 indefinitely.
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'}
Related Topics: Math
Solution 1.
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class Solution { public boolean isRobotBounded(String instructions) { if (instructions == null || instructions.length() == 0) return true; int[][] directions = { {0, 1}, {1, 0}, {0, -1}, {-1, 0} }; int directionIndex = 0; int xPosition = 0, yPosition = 0; int length = instructions.length(); for (int i = 0; i < length; i++) { char instruction = instructions.charAt(i); if (instruction == 'G') { int[] direction = directions[directionIndex]; xPosition += direction[0]; yPosition += direction[1]; } else if (instruction == 'L') directionIndex = (directionIndex + 3) % 4; else if (instruction == 'R') directionIndex = (directionIndex + 1) % 4; } return xPosition == 0 && yPosition == 0 || directionIndex != 0; } } ############ class Solution { public boolean isRobotBounded(String instructions) { int[] direction = new int[4]; int cur = 0; for (char c : instructions.toCharArray()) { if (c == 'L') { cur = (cur + 1) % 4; } else if (c == 'R') { cur = (cur + 3) % 4; } else { ++direction[cur]; } } return cur != 0 || (direction[0] == direction[2] && direction[1] == direction[3]); } }
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// OJ: https://leetcode.com/problems/robot-bounded-in-circle/ // Time: O(N) // Space: O(1) class Solution { public: bool isRobotBounded(string instructions) { int x = 0, y = 0, d = 0, dir[4][2] = { {0,1},{-1,0},{0,-1},{1,0} }; for (int i = 0; i < 4; ++i) { for (char c : instructions) { if (c == 'G') x += dir[d][0], y += dir[d][1]; else if (c == 'L') d = (d + 1) % 4; else d = (d + 3) % 4; } if (x == 0 && y == 0) return true; } return false; } };
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class Solution: def isRobotBounded(self, instructions: str) -> bool: cur, direction = 0, [0] * 4 for ins in instructions: if ins == 'L': cur = (cur + 1) % 4 elif ins == 'R': cur = (cur + 3) % 4 else: direction[cur] += 1 return cur != 0 or ( direction[0] == direction[2] and direction[1] == direction[3] ) ############ class Solution(object): def isRobotBounded(self, instructions): """ :type instructions: str :rtype: bool """ dirs = [(0, 1), (-1, 0), (0, -1), (1, 0)] x, y = 0, 0 curd = 0 for i in instructions: if i == "G": x += dirs[curd][0] y += dirs[curd % 4][1] elif i == "L": curd = (curd + 1) % 4 elif i == "R": curd = (curd - 1) % 4 return (x == 0 and y == 0) or curd != 0
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func isRobotBounded(instructions string) bool { direction := make([]int, 4) cur := 0 for _, ins := range instructions { if ins == 'L' { cur = (cur + 1) % 4 } else if ins == 'R' { cur = (cur + 3) % 4 } else { direction[cur]++ } } return cur != 0 || (direction[0] == direction[2] && direction[1] == direction[3]) }
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function isRobotBounded(instructions: string): boolean { const dist: number[] = new Array(4).fill(0); let k = 0; for (const c of instructions) { if (c === 'L') { k = (k + 1) % 4; } else if (c === 'R') { k = (k + 3) % 4; } else { ++dist[k]; } } return (dist[0] === dist[2] && dist[1] === dist[3]) || k !== 0; }