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1041. Robot Bounded In Circle
On an infinite plane, a robot initially stands at (0, 0) and faces north. The robot can receive one of three instructions:
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"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:
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1 ⇐ instructions.length ⇐ 100 -
instructions[i]is in{'G', 'L', 'R'}
思路分析
| 这道题不是要看路径是否相交。 |
这道题的关键点事找出不存在环路的条件：执行一遍指令后，机器人必须不在原点且方向继续朝北：
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如果在原点，无论执行多少遍，结果都会在原点。
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如果方向不是初始化方向，那么多次执行后，就会相互抵消，形成环路。
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-27 16:10:21
*/
public boolean isRobotBounded(String instructions) {
int x = 0, y = 0;
int dx = 0, dy = 1;
for (int i = 0; i < instructions.length(); i++) {
char step = instructions.charAt(i);
if (step == 'G') {
x += dx;
y += dy;
} else if (step == 'R') {
int temp = dx;
dx = dy;
dy = -temp;
} else if (step == 'L') {
int temp = dx;
dx = -dy;
dy = temp;
}
}
return !(dx == 0 && dy == 1) || (x == 0 && y == 0);
}