LeetCode #657 Robot Return to Origin 机器人能否返回原点

657 Robot Return to Origin 机器人能否返回原点

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:

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.

题目描述:
在二维平面上，有一个机器人从原点 (0, 0) 开始。给出它的移动顺序，判断这个机器人在完成移动后是否在 (0, 0) 处结束。

移动顺序由字符串表示。字符 move[i] 表示其第 i 次移动。机器人的有效动作有 R（右），L（左），U（上）和 D（下）。如果机器人在完成所有动作后返回原点，则返回 true。否则，返回 false。

注意：
机器人“面朝”的方向无关紧要。 “R” 将始终使机器人向右移动一次，“L” 将始终向左移动等。此外，假设每次移动机器人的移动幅度相同。

示例 :

示例 1:

输入: "UD"
输出: true
解释：机器人向上移动一次，然后向下移动一次。所有动作都具有相同的幅度，因此它最终回到它开始的原点。因此，我们返回 true。
示例 2:

输入: "LL"
输出: false
解释：机器人向左移动两次。它最终位于原点的左侧，距原点有两次 “移动” 的距离。我们返回 false，因为它在移动结束时没有返回原点。

思路:

遍历字符串, 比较 U和D以及 L和 R的出现次数
时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

class Solution 
{
public:
    bool judgeCircle(string moves) 
    {
        int count[26] = {0};
        for (char c : moves) ++count[c - 'A'];
        return count['L' - 'A'] == count['R' - 'A'] and count['U' - 'A'] == count['D' - 'A'];
    }
};

Java:

class Solution {
    public boolean judgeCircle(String moves) {
        int[] count = new int[26];
        for (char c : moves.toCharArray()) count[c - 'A']++;
        return count['L' - 'A'] == count['R' - 'A'] && count['U' - 'A'] == count['D' - 'A'];
    }
}

Python:

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