【LeetCode】657. Robot Return to Origin 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • • 复数求和
    • counter统计次数
  • 相似题目
  • 参考资料
  • 日期

题目地址：https://leetcode.com/problems/robot-return-to-origin/description/

题目描述

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.

题目大意

输入是机器人移动的方向，每次移动一步。问最终是否出现在出发位置。机器人的面向和移动方向是不想关的。

解题方法

复数求和

Python支持复数运算的，所以指定不同的方向是不同的实数数字就行了。复数的标记是j。最后求和判断是不是0就是到了原点。

时间复杂度是O(n)，空间复杂度是O(1).

class Solution:
    def judgeCircle(self, moves):
        """
        :type moves: str
        :rtype: bool
        """
        directs = {'L':-1, 'R':1, 'U':1j, 'D':-1j}
        return 0 == sum(directs[move] for move in moves)

counter统计次数

显而易见，回到原点的要求是向上走的次数和向下走的次数相等，并且向左和向右走的次数相等。直接字符串统计判断是否哦相等，很快就写出来。

时间复杂度是O(n)，空间复杂度是O(1).

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

上面的代码也可以直接使用4个变量统计次数，但是时间并没有明显提升。

class Solution:
    def judgeCircle(self, moves):
        """
        :type moves: str
        :rtype: bool
        """
        u = d = l = r = 0
        for move in moves:
            if move == 'U':
                u += 1
            elif move == "D":
                d += 1
            elif move == 'L':
                l += 1
            elif move == 'R':
                r += 1
        return u == d and l == r

相似题目

参考资料

日期

2018 年 11 月 2 日 —— 浑浑噩噩的一天