[Swift]LeetCode1041. 困于环中的机器人 | 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:

  • "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 indefinetely.

Example 3:

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

Note:

  1. 1 <= instructions.length <= 100
  2. instructions[i] is in {'G', 'L', 'R'

在无限的平面上,机器人最初位于 (0, 0) 处,面朝北方。机器人可以接受下列三条指令之一:

  • "G":直走 1 个单位
  • "L":左转 90 度
  • "R":右转 90 度

机器人按顺序执行指令 instructions,并一直重复它们。

只有在平面中存在环使得机器人永远无法离开时,返回 true。否则,返回 false

示例 1:

输入:"GGLLGG"
输出:true
解释:
机器人从 (0,0) 移动到 (0,2),转 180 度,然后回到 (0,0)。
重复这些指令,机器人将保持在以原点为中心,2 为半径的环中进行移动。

示例 2:

输入:"GG"
输出:false
解释:
机器人无限向北移动。

示例 3:

输入:"GL"
输出:true
解释:
机器人按 (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ... 进行移动。

提示:

  1. 1 <= instructions.length <= 100
  2. instructions[i] 在 {'G', 'L', 'R'} 中

Runtime: 8 ms
Memory Usage: 20.6 MB
 1 class Solution {
 2     func isRobotBounded(_ instructions: String) -> Bool {
 3         var arr:[Character] = Array(instructions)
 4         var left:Int = 0
 5         var right:Int = 0
 6         for c in arr
 7         {
 8             if c == "L"
 9             {
10                 left += 1
11                 
12             }
13             if c == "R"
14             {
15                 right += 1
16             }
17         }
18         if left == 0 && right == 0
19         {
20             return false
21         }
22         if left != 0 && right == left
23         {
24             return false
25         }
26         return true
27     }
28 }

猜你喜欢

转载自www.cnblogs.com/strengthen/p/10851797.html