LeetCode-Robot Return to Origin

最新推荐文章于 2020-02-14 20:59:08 发布

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本文介绍了一种判断机器人是否能从原点出发并按给定的指令序列返回原点的方法。通过跟踪水平和垂直方向的位移，可以简单有效地判断机器人最终位置。

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Description：
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

题意：有一个处在原点（0,0）的机器人，给定一系列的指令，包括’U’，’D’，’L’，’R’；判断最后机器人是否能够返回原点；

解法：现在假定机器人处在二维平面的原点上，要想使其最后返回原点，那么机器人向左走动的距离应当和向右走动的距离相等，同理，向上走动的距离和向下走动的距离相等；因此，我们只需要两个变量记录水平方向和垂直方向是否最后处在原点即可；

class Solution {
    public boolean judgeCircle(String moves) {
        int hDirection = 0; //horizontal direction
        int vDirection = 0; //vertical direction
        for (int i = 0; i < moves.length(); i++) {
            switch (moves.charAt(i)) {
                case 'U':
                    vDirection++;
                    break;
                case 'D':
                    vDirection--;
                    break;
                case 'L':
                    hDirection--;
                    break;
                case 'R':
                    hDirection++;
                default:
                    break;
            }
        }
        return vDirection == 0 && hDirection == 0 ? true : false;
    }
}