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.

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题目大意：

  机器人处于原点(0,0),给定一个字符串数组moves，表示机器人的运动轨迹，UDLR四个字符分别表示上下左右四个方向，判断机器人运动结束后是否处于原点。

解题思路：

  设置两个变量x、y分别表示机器人运动后的XY轴坐标位置。最后判断x、y是否都为0。

解题代码：

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);
    }
};