 DAY 21
0

Question

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

Example1

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.

Example2

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.

Example3

Input: moves = "RRDD"
Output: false

Example4

Input: moves = "LDRRLRUULR"
Output: false

Constraints

• 1 <= moves.length <= 2 * 10^4
• moves only contains the characters 'U', 'D', 'L' and 'R'.

解題

Think

• 分x, y座標，讀到'R'=> +1, 'L'=> -1, 'U'=> 1, 'D'=> -1，最後判斷座標是不是有任何一個是非0的值。
• C的程式睡醒再補上。

Code

Python

class Solution:
def judgeCircle(self, moves: str) -> bool:
dict_moves = {'R':1, 'L':-1, 'U':1, 'D':-1}
loc = [0, 0]

for index in range(len(moves)):
if moves[index] == 'R' or moves[index] == 'L':
loc += dict_moves[moves[index]]

elif moves[index] == 'U' or moves[index] == 'D':
loc += dict_moves[moves[index]]

if loc != 0 or loc != 0:
return False
else:
return True

C

bool judgeCircle(char * moves){
int loc = {0};

for (int index=0 ; index<strlen(moves) ; index++){
if (moves[index] == 'R'){
loc++;
} else if (moves[index] == 'L'){
loc--;
} else if (moves[index] == 'U'){
loc++;
} else if (moves[index] == 'D'){
loc--;
}
}

if (loc != 0 || loc != 0){
return false;
} else {
return true;
}

}

Result

• Python

• • C

• 30天 Leetcode解題之路30