leetcode 657. Robot Return to Origi(C++)

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.
テーマ:
ロボットは原点(0,0)にあり、1つの文字列配列movesが与えられ、ロボットの動き軌跡を表し、UDLRの4文字がそれぞれ の4方向を表し、ロボットの動きが終了した後に原点にあるか否かを判断する.
問題解決の考え方:
2つの変数x、yがそれぞれロボットの動きを表すXY軸座標位置を設定します.最後に、x、yがいずれも0であるか否かを判断する.
解題コード:

class Solution {
public:
    bool judgeCircle(string moves) {
        int x = 0, y = 0;
        for(char ch : moves){
            switch(ch){
                case 'U': ++y; break;
                case 'D': --y; break;
                case 'L': --x; break;
                case 'R': ++x; break;
                default : break;
            }
        }
        return (x == 0 && y == 0);
    }
};