---
title: "657. Robot Return to Origin"
created: 2025-12-19
---
# 657. Robot Return to Origin

## 题目 [**657. Robot Return to Origin**](https://leetcode.com/problems/robot-return-to-origin/)

![[image-d1ba545a.png]]

## 思路分析

![[image-0a34d238.png]]

```java
class Solution {
    public boolean judgeCircle(String moves) {
        StringBuilder stack = new StringBuilder();
        for(int i=0;i<moves.length();i++){
            char cur = moves.charAt(i);
            if(stack.length()>0){
                char top=stack.charAt(stack.length()-1);
                if(isPair(top,cur)){
                    stack.deleteCharAt(stack.length()-1);
                    continue;
                }
            }
            stack.append(cur);
        }
        return stack.length()==0;
    }

    private boolean isPair(char a,char b){
        if (a == 'U' && b == 'D') return true;
        if (a == 'D' && b == 'U') return true;
        if (a == 'L' && b == 'R') return true;
        if (a == 'R' && b == 'L') return true;
        return false;
    }
}
```

但其实发现会错 聪明反被聪明误了

栈的一个核心特性是**必须消除相邻（或经消除后相邻）的元素**。但这道题是二维平面的移动，`X轴` 的移动和 `Y轴` 的移动是**互不干扰**的，中间隔着别的方向也能抵消。

其实只需要看数量++ –-最后等不等于0即可

## 代码实现

```java
class Solution {
    public boolean judgeCircle(String moves) {
        int[] cnt = new int[26];

        for(char c:moves.toCharArray()){
            cnt[c-'A']++;
        }

        return  cnt['U'-'A']==cnt['D'-'A'] &&
                cnt['L'-'A']==cnt['R'-'A'];
    }
}
```

```java
class Solution {
    public boolean judgeCircle(String moves) {
        int x = 0, y = 0;
        for (int i = 0; i < moves.length(); i++) {
            switch (moves.charAt(i)) {
                case 'U': y++; break;
                case 'D': y--; break;
                case 'L': x--; break;
                case 'R': x++; break;
            }
        }
        return x == 0 && y == 0;
    }
}
```

## 同类题型

## 视频讲解