以下以 Leetcode 657. Robot Return to Origin 為例子，
程式的解答只有十行程式碼，
其實大部份的程式人員都寫的出來，但是，程式人員在英文閱讀時，一定會卡關。所以，我特別寫出來這篇文章。
Leetcode 657題目英文敘述的長度，快接近2篇 A4 大小。

每個程式設計師都會寫leetcode比較困難的是在於閱讀英文題目
上面這句話，是我進行真人面試20次後的心得。

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

其中一種，閱讀策略:

1. 先看「Example 1:」「Output: false」
2. 看最後的「Constraints:」
3. 看題目，從英文文章後面看始看「Note: The way that the robot」「Return true if the robot」「You are given a string moves」

Leetcode 657. Robot Return to Origin 題目
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 歐育溙
version: 2023112701