Robot Bounded In Circle (M)
題目
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 <= 100instructions[i]is in{'G', 'L', 'R'}
題意
給定一連串機器人行動的指令，判斷機器人是否能保持在一個圓圈范圍內活動，
思路
要讓機器人保持在圓圈范圍內活動，就要保持機器人會周期性的通過原點，當一個周期的指令執行完畢后，機器人會發生一個坐標的變化和一個指向方向的變化：如果結束方向指向北方，那么只有當結束位置也在原點時才能滿足條件；如果結束方向指向南方，那么經過兩個周期的指令一定能回到原點；如果結束方向指向西方或者東方，那么經過四個周期的指令一定能回到原點，因此只要對結束方向為北方這一種情況進行判斷，
代碼實作
Java
class Solution {
public boolean isRobotBounded(String instructions) {
int diffX = 0, diffY = 0;
int dir = 0;
for (char c : instructions.toCharArray()) {
if (c == 'G') {
diffX += dir == 1 ? -1 : dir == 3 ? 1 : 0;
diffY += dir == 0 ? 1 : dir == 2 ? -1 : 0;
} else if (c == 'L') {
dir = (dir + 3) % 4;
} else {
dir = (dir + 1) % 4;
}
}
return dir == 0 ? diffX == 0 && diffY == 0 : true;
}
}
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標籤：其他