题目

机器人从（0,0）出发，初始化向北

三种指令 G当前方向+1步，L左转90度，R右转90度

问指令结束后是否成圆圈（只有可能回原点时才成圈）

返回true/false

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 degrees 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: instructions = "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: instructions = "GG"
Output: false
Explanation: The robot moves north indefinitely.

Example 3:

Input: instructions = "GL"
Output: true
Explanation: The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ...

 

思路

分类讨论

1、转了一圈回来

2、原地打转没前进

 

找规律，【原，右，x，左】

第i次right命令=（i+1）%4

第i次left命令=（i+3）%4

 

 

 

 

代码

细节踩坑，注意

if..if..else

if..else if..else

的区别，测试案例"GLRLLGLL"

 

注意只转圈未前进的情况（i>0）

测试案例RL

class Solution {
    public boolean isRobotBounded(String instructions) {
        int x=0,y=0,help[][]={{0,1},{1,0},{0,-1},{-1,0}};//原、右、x、左
        int i=0;
        for(int j=0;j<instructions.length();++j){
            if(instructions.charAt(j)=='L'){
                i=(i+3)%4;
            }
            else if(instructions.charAt(j)=='R'){
                i=(i+1)%4;
            }
            else{
                x+=help[i][0];
                y+=help[i][1];
            }
        }
        return x==0&&y==0||i>0;
    }
}