每个程式设计师都会写leetcode比较困难的是在于阅读英文题目 leetcode 中文

以下以 Leetcode 657. Robot Return to Origin 为例子，
程式的解答只有十行程式码，
其实大部份的程式人员都写的出来，但是，程式人员在英文阅读时，一定会卡关。所以，我特别写出来这篇文章。
Leetcode 657题目英文叙述的长度，快接近2篇 A4 大小。

每个程式设计师都会写leetcode比较困难的是在于阅读英文题目
上面这句话，是我进行真人面试20次后的心得。

以下是 Python 解答

class Solution(object):    def judgeCircle(self, moves):        x = y = 0        for move in moves:            if move == 'U': y -= 1            elif move == 'D': y += 1            elif move == 'L': x -= 1            elif move == 'R': x += 1        return x == y == 0

其中一种，阅读策略:

先看「Example 1:」「Output: false」看最后的「Constraints:」看题目，从英文文章后面看始看「Note: The way that the robot」「Return true if the robot」「You are given a string moves」

Leetcode 657. Robot Return to Origin 题目
There is a robot starting at the 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.

You are given a string moves that represents the move sequence of the robot where moves[i] represents its ith move. Valid moves are 'R' (right), 'L' (left), 'U' (up), and 'D' (down).

Return true if the robot returns to the origin after it finishes all of its moves, or false otherwise.

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: moves = "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: moves = "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.

Constraints:

1 <= moves.length <= 2 * 104
moves only contains the characters 'U', 'D', 'L' and 'R'.

文章作者
Billour Ou 欧育溙
version: 2023112701