LeetCode #657 RobotReturnToOrigin. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Nice performance
Runtime: 16 ms, faster than 94.48% of C++ online submissions for Robot Return to Origin.
Memory Usage: 10.3 MB, less than 52.06% of C++ online submissions for Robot Return to Origin.
LeetCode #657
Q.
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.
로봇 움직임의 시작점 (0, 0) 이 2차원에 있다. 일련의 움직임에서, 로봇이 움직임을 마치고 다시 (0, 0)으로 오는지 판단해봐라.
움직임들은 문자열로 나타나는데, moves[i] 는 움직임 문자이다. R은 오른쪽, L 은 왼쪽, U 는 위쪽, D 는 아랫쪽이다. 만약에 로봇이 모든 움직임 후에 원점으로 돌아오면 true 이고 아니면 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.
로봇은 움직임마다 항상 같은 거리를 움직인다.
e.g.
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.
Process
// Process
//1. Input moves string
//2. Iterate from begin to the end
// 2.1. Check if it's U, R, D, or L
// 2.1.1. If it's U -> ++ y
// 2.1.2. if else it's R -> ++ x
// 2.1.3. if else it's D -> -- y
// 2.1.4. if else it's L -> -- x
//3. If point x, y are 0 -> answer is true
//4. Return answer
// 처리과정
//1. 움직임 문자열을 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1. U, R, D, L 중 어떤 것인지 확인한다.
// 2.1.1. U 이면 ++y
// 2.1.2. R 이면 ++x
// 2.1.3. D 이면 --y
// 2.1.4. L 이면 --x
//3. 포인트 x, y 둘다 0 이면 true
//4. 결과 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
bool judgeCircle(string moves) {
bool answer = false;
if (moves.size() > 0) {
int point[2] = { 0, 0 };
for (int i = 0; i < moves.size(); ++i) {
switch (moves[i])
{
case 'R':
++point[0]; break;
case 'U':
++point[1]; break;
case 'L':
--point[0]; break;
case 'D':
--point[1]; break;
default:
break;
}
}
if (point[0] == 0 && point[1] == 0) {
answer = true;
}
}
return answer;
}
};
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