LeetCode 657. Robot Return to Origin

秒速五厘米 2021-12-14 13:55 128阅读 0赞

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.

**题目描述：**大概意思是求一个点经过若干次移动能否回到原来的位置。

**题目分析：**我们只需要去判断两个条件：

*  向左移动的次数和向右移动的次数是否相等
 *  向上移动的次数是否和向下移动的次数相等

于是我们只需要计算出向左、向右、向上、向下分别走了多少步就行了。

`python` 代码：

class Solution(object): def judgeCircle(self, moves): """ :type moves: str :rtype: bool """ moves_length = len(moves) L_count = 0 R_count = 0 U_count = 0 D_count = 0 for i in range(moves_length): if moves[i] == 'L': L_count = L_count + 1 elif moves[i] == 'R': R_count = R_count + 1 elif moves[i] == 'U': U_count = U_count + 1 elif moves[i] == 'D': D_count = D_count + 1 if L_count == R_count and U_count == D_count: return True else: return False

`C++` 代码：

class Solution { public: bool judgeCircle(string moves) { int moves_length = moves.length(); int L_count = 0; int R_count = 0; int U_count = 0; int D_count = 0; for(int i = 0; i < moves_length; i++){ if(moves[i] == 'L'){ L_count++; } else if(moves[i] == 'R'){ R_count++; } else if(moves[i] == 'U'){ U_count++; } else if(moves[i] == 'D'){ D_count++; } } if(L_count == R_count && U_count == D_count){ return true; } return false; } };

转载于:https://www.cnblogs.com/ECJTUACM-873284962/p/10234583.html