README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0600-0699/0657.Robot%20Return%20to%20Origin/README_EN.md	String Simulation

657. Robot Return to Origin

中文文档

Description

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'.

Solutions

Solution 1: Maintain Coordinates

We can maintain a coordinate $(x, y)$ to represent the robot's movement in the horizontal and vertical directions.

Traverse the string $\textit{moves}$ and update the coordinate $(x, y)$ based on the current character:

• If the current character is 'U', then $y$ increases by $1$;
• If the current character is `'D'$, then $y$ decreases by $1$;
• If the current character is `'L'$, then $x$ decreases by $1$;
• If the current character is `'R'$, then $x$ increases by $1$.

Finally, check if both $x$ and $y$ are $0$.

The time complexity is $O(n)$, where $n$ is the length of the string $\textit{moves}$. The space complexity is $O(1)$.

Python3

class Solution:
    def judgeCircle(self, moves: str) -> bool:
        x = y = 0
        for c in moves:
            match c:
                case "U":
                    y += 1
                case "D":
                    y -= 1
                case "L":
                    x -= 1
                case "R":
                    x += 1
        return x == 0 and y == 0

Java

class Solution {
    public boolean judgeCircle(String moves) {
        int x = 0, y = 0;
        for (char c : moves.toCharArray()) {
            switch (c) {
                case 'U' -> y++;
                case 'D' -> y--;
                case 'L' -> x--;
                case 'R' -> x++;
            }
        }
        return x == 0 && y == 0;
    }
}

C++

class Solution {
public:
    bool judgeCircle(string moves) {
        int x = 0, y = 0;
        for (char c : moves) {
            switch (c) {
            case 'U': y++; break;
            case 'D': y--; break;
            case 'L': x--; break;
            case 'R': x++; break;
            }
        }
        return x == 0 && y == 0;
    }
};

Go

func judgeCircle(moves string) bool {
	x, y := 0, 0
	for _, c := range moves {
		switch c {
		case 'U':
			y++
		case 'D':
			y--
		case 'L':
			x--
		case 'R':
			x++
		}
	}
	return x == 0 && y == 0
}

TypeScript

function judgeCircle(moves: string): boolean {
    let [x, y] = [0, 0];
    for (const c of moves) {
        if (c === 'U') {
            y++;
        } else if (c === 'D') {
            y--;
        } else if (c === 'L') {
            x--;
        } else {
            x++;
        }
    }
    return x === 0 && y === 0;
}

JavaScript

/**
 * @param {string} moves
 * @return {boolean}
 */
var judgeCircle = function (moves) {
    let [x, y] = [0, 0];
    for (const c of moves) {
        if (c === 'U') {
            y++;
        } else if (c === 'D') {
            y--;
        } else if (c === 'L') {
            x--;
        } else {
            x++;
        }
    }
    return x === 0 && y === 0;
};