LeetCode 657 - Robot Return to Origin

https://leetcode.com/problems/robot-return-to-origin/

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.

https://leetcode.com/articles/judge-route-circle/

We can simulate the position of the robot after each command.

Algorithm

Initially, the robot is at (x, y) = (0, 0). If the move is 'U', the robot goes to (x, y-1); if the move is 'R', the robot goes to (x, y) = (x+1, y), and so on.

public boolean judgeCircle(String moves) {
    int x = 0, y = 0;
    for (char move: moves.toCharArray()) {
        if (move == 'U') y--;
        else if (move == 'D') y++;
        else if (move == 'L') x--;
        else if (move == 'R') x++;
    }
    return x == 0 && y == 0;
}

https://leetcode.com/problems/robot-return-to-origin/discuss/106316/C%2B%2B-Java-Clean-Code

    public boolean judgeCircle(String moves) {
        int x = 0;
        int y = 0;
        for (char ch : moves.toCharArray()) {
            if (ch == 'U') y++;
            else if (ch == 'D') y--;
            else if (ch == 'R') x++;
            else if (ch == 'L') x--;
        }
        return x == 0 && y == 0;
    }

https://leetcode.com/problems/robot-return-to-origin/discuss/106342/Easy-2-lines-Java

Easy solution using split.(It needs spaces from front and behind to be calculated correctly):

 public boolean judgeCircle(String moves) {
        moves=" " + moves + " ";
        return moves.split("L").length==moves.split("R").length && moves.split("U").length == moves.split("D").length;
    }

public boolean judgeCircle(String moves) {
        int i = 0;
        int j = 0;
        char[] chars = moves.toCharArray();
        for (char ch : chars) {
                if (ch == 'U') {
                        i += 1;
                } else if (ch == 'D') {
                        i -= 1;
                } else if (ch == 'R') {
                        j += 1;
                } else if (ch == 'L') {
                        j -= 1;
                }
        }
        return i == 0 && j == 0;
}

https://leetcode.com/problems/robot-return-to-origin/discuss/194765/Java-one-liner

public boolean judgeCircle(String moves) {
        
                return moves.replace("L", "").length() == moves.replace("R", "").length()
            && moves.replace("U", "").length() == moves.replace("D", "").length(); 
    }

http://www.zhaojun.im/leetcode-657/

其实就是给一个字符串, 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。

这道题很简单，只需要假设当前节点是 0, 0，定义两个变量, i 和 j，默认值都为 0，每当向上 i + 1，向下 i - 1，向右 j + 1，向左 j - 1。最终只需要判断 i 和 j 是否都等于 0 即可。