657. Robot Return to Origin

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本文介绍了一种算法，用于判断二维平面上的机器人在完成一系列上（U）、下（D）、左（L）、右（R）移动后是否能回到起点（0,0）。通过计算移动次数或使用复数进行方向映射，可以简洁地实现这一目标。

657. Robot Return to Origin

Easy

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.

给一个字符串, 每个字符包含 “U”、”D”、”L”、”R”, 分别表示上下左右, 表示机器人向这个位置走一步, 判断最终是否机器人是否还在原来的位置。

Python

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        i = 0
        j = 0
        for ch in moves:
            if ch == 'U':
                i += 1
            elif ch =='D':
                i -= 1
            elif ch == 'L':
                j += 1
            elif ch == 'R':
                j -= 1
        return i == 0 and j == 0
      
class Solution:
    def judgeCircle(self, moves: str) -> bool:
        directs = {'L':-1, 'R':1, 'U':1j, 'D':-1j}
        return 0 == sum(directs[move] for move in moves)

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