A robot on an infinite grid starts at point (0, 0) and faces north. The robot can receive one of three possible types of commands:

• -2: turn left 90 degrees
• -1: turn right 90 degrees
• 1 <= x <= 9: move forward x units

Some of the grid squares are obstacles.

The i-th obstacle is at grid point (obstacles[i][0], obstacles[i][1])

If the robot would try to move onto them, the robot stays on the previous grid square instead (but still continues following the rest of the route.)

Return the square of the maximum Euclidean distance that the robot will be from the origin.

Example 1:

Input: commands = [4,-1,3], obstacles = []
Output: 25
Explanation: robot will go to (3, 4)
 

Example 2:

Input: commands = [4,-1,4,-2,4], obstacles = [[2,4]]
Output: 65
Explanation: robot will be stuck at (1, 4) before turning left and going to (1, 8)

Note:

1. 0 <= commands.length <= 10000
2. 0 <= obstacles.length <= 10000
3. -30000 <= obstacle[i][0] <= 30000
4. -30000 <= obstacle[i][1] <= 30000
5. The answer is guaranteed to be less than 2 ^ 31.

解法
傳統的模擬方法即可，把障礙物全都放到set裡面，一步一步走。這樣子即容易實現又不會出錯。一開始的思路是判斷當前點和下一點之間有沒有障礙物，這種方法是錯誤的。

class Solution {
public:
    int robotSim(vector<int>& cmd, vector<vector<int>>& obs) {
        int dir[][2] = {{0,1},{1,0},{0,-1},{-1, 0}};
        int x=0,y=0;
	    int curdir = 0;
        int n = cmd.size(), m=obs.size();
        int ans = 0;
        set<pair<int,int>> block;
        for(int i=0;i < m ;i++) {
            block.insert({obs[i][0], obs[i][1]});
        }
        for(int i=0;i<n;i++) {
            if(cmd[i]==-1) {
                curdir = (curdir+1)%4;
            }
            else if(cmd[i]==-2) {
                curdir = (curdir+3)%4;
            }
            else{
                int step = cmd[i];
                int x1 =x , y1 = y;
                while(step--) {
                    x1 += dir[curdir][0],y1 += dir[curdir][1];
                    if (block.find({x1, y1}) != block.end())  {
                        break;
                    }
                    x=x1,y=y1;
                }
                ans = max(ans, x*x+y*y);
            }
	    }
        return ans;
    }
};