leetcode 874. 模拟行走机器人(Walking Robot Simulation)
题目描述:
机器人在一个无限大小的网格上行走,从点 (0, 0) 处开始出发,面向北方。该机器人可以接收以下三种类型的命令:
-2
:向左转 90 度-1
:向右转 90 度1 <= x <= 9
:向前移动x
个单位长度
在网格上有一些格子被视为障碍物。
第 i
个障碍物位于网格点 (obstacles[i][0], obstacles[i][1])
如果机器人试图走到障碍物上方,那么它将停留在障碍物的前一个网格方块上,但仍然可以继续该路线的其余部分。
返回从原点到机器人的最大欧式距离的平方。
示例 1:
输入: commands = [4,-1,3], obstacles = []
输出: 25
解释: 机器人将会到达 (3, 4)
示例 2:
输入: commands = [4,-1,4,-2,4], obstacles = [[2,4]]
输出: 65
解释: 机器人在左转走到 (1, 8) 之前将被困在 (1, 4) 处
提示:
0 <= commands.length <= 10000
0 <= obstacles.length <= 10000
-30000 <= obstacle[i][0] <= 30000
-30000 <= obstacle[i][1] <= 30000
- 答案保证小于
2 ^ 31
解法:
# define PR pair<long long, long long>
class Solution {
public:
bool move(int& _x, int& _y, bool up, bool down, bool left, bool right, unordered_set<long long>& st){
int x = _x, y = _y;
if(up){
y++;
}else if(down){
y--;
}else if(left){
x--;
}else{
x++;
}
long long pos = parse({x, y});
if(st.find(pos) == st.end()){
_x = x;
_y = y;
return true;
}else{
return false;
}
}
long long parse(PR pr){
long long x = pr.first, y = pr.second;
return (x - 90000)*90000 + (y - 90000);
}
int robotSim(vector<int>& commands, vector<vector<int>>& obstacles) {
unordered_set<long long> st;
for(vector<int> obs : obstacles){
st.insert(parse({obs[0], obs[1]}));
}
int x = 0, y = 0;
int res = 0;
bool up = true;
bool down = false, left = false, right = false;
for(int cmd : commands){
if(cmd == -1){
bool tmp = right;
right = up;
up = left;
left = down;
down = tmp;
}else if(cmd == -2){
bool tmp = up;
up = right;
right = down;
down = left;
left = tmp;
}else{
for(int i = 0; i < cmd; i++){
if(!move(x, y, up, down, left, right, st)){
break;
}else{
// cout<<x<<", "<<y<<endl;
res = max(res, x*x + y*y);
}
}
}
}
return res;
}
};