LeetCode 657 : Robot Return to Origin

题目地址：Robot Return to Origin - LeetCode

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Acceptance:70.8%
Difficulty:Easy

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There is a robot starting at position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves.

The move sequence is represented by a string, and the character moves[i] represents its ith move. Valid moves are R (right), L (left), U (up), and D (down). If the robot returns to the origin after it finishes all of its moves, return true. Otherwise, return false.

Note: The way that the robot is “facing” is irrelevant. “R” will always make the robot move to the right once, “L” will always make it move left, etc. Also, assume that the magnitude of the robot’s movement is the same for each move.

Example 1:

Input: "UD"
Output: true 
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.

Example 2:

Input: "LL"
Output: false
Explanation: The robot moves left twice. It ends up two "moves" to the left of the origin. We return false because it is not at the origin at the end of its moves.

题目的意思是判断一个机器人在上下左右移动后能否回到原点，只要计算左移的个数和右移的个数相同，上下的个数相同就行。
Python代码：

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        return moves.count('L')==moves.count('R') and moves.count('U')==moves.count('D')

结果如下，还是很快的：

Runtime: 40 ms, faster than 91.13% of Python3 online submissions for Robot Return to Origin.
Memory Usage: 13.2 MB, less than 5.32% of Python3 online submissions for Robot Return to Origin.

java代码如下：

class Solution {
    
    public boolean judgeCircle(String moves) {
         int real=0;  
     	int image=0; 
        for (char move: moves.toCharArray()) {
            if (move == 'U') real--;
            else if (move == 'D') real++;
            else if (move == 'L') image--;
            else if (move == 'R') image++;
        }
        return real == 0 && image == 0;
    }
}