描述

最初，机器人位于(0, 0)处。给定一系列动作，判断该机器人的移动轨迹是否是一个环，这意味着它最终会回到原来的位置。

移动的顺序由字符串表示。每个动作都由一个字符表示。有效的机器人移动是R（右），L（左），U（上）和D（下）。输出应该为true或false，表示机器人是否回到原点。

样例

样例1:

输入: "UD"
输出: true

样例2:

输入: "LL"
输出: false

编程语言：Java

解题思想：

1.以坐标的形式来移动位置和判断位置。

2.因为只有四种走法，可以使用switch进行判断。

时间复杂度：O(n)

IDE代码如下：

package LintCode_Easy;
 
public class _1104{
 
	public static void main(String[] args) {
		String moves="LL";
		int xplace = 0,yplace=0;
		for(int i=0;i<moves.length();i++)
		{
			switch (moves.charAt(i)) {
			case 'U':yplace++;break;
			case 'D':yplace--;break;
			case 'L':xplace--;break;
			case 'R':xplace++;break;
			default:
				break;
			}
		}
		boolean placejudge=(xplace==0 && yplace==0)?true:false;
		System.out.println(placejudge);
	}
}

LintCode编辑器代码如下：

public class Solution {
    public boolean judgeCircle(String moves) {
       int xplace = 0,yplace=0;
		for(int i=0;i<moves.length();i++)
		{
			switch (moves.charAt(i)) {
			case 'U':yplace++;break;
			case 'D':yplace--;break;
			case 'L':xplace--;break;
			case 'R':xplace++;break;
			default:
				break;
			}
		}
		boolean placejudge=(xplace==0 && yplace==0)?true:false;
		return placejudge;
    }
}

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LintCode_easy:机器人能否返回原点

描述

样例