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.

題目大意：

??機器人處于原點(0,0),給定一個字符串?dāng)?shù)組moves，表示機器人的運動軌跡，UDLR四個字符分別表示上下左右四個方向，判斷機器人運動結(jié)束后是否處于原點。

解題思路：

??設(shè)置兩個變量x、y分別表示機器人運動后的XY軸坐標(biāo)位置。最后判斷x、y是否都為0。

解題代碼：

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);
    }
};