657. Robot Return to Origin（python+cpp）

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题目：

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 rigin. We return false because it is not at the origin at the end of its moves.

解释：
判断机器人是否回到原位，直接数数，看能否抵消即可
python代码：

class Solution(object):
    def judgeCircle(self, moves):
        """
        :type moves: str
        :rtype: bool
        """
        count_U=moves.count('U')
        count_D=moves.count('D')
        count_L=moves.count('L')
        count_R=moves.count('R')
        if(count_U==count_D and count_L==count_R):
            return True
        else:
            return False  

c++代码：

#include <algorithm>
class Solution {
public:
    bool judgeCircle(string moves) {
        int count_U=count(moves.begin(),moves.end(),'U');
        int count_D=count(moves.begin(),moves.end(),'D');
        int count_L=count(moves.begin(),moves.end(),'L');
        int count_R=count(moves.begin(),moves.end(),'R');
        if (count_U==count_D and count_R == count_L )
            return true;
        else
            return false;
    }
};

总结：
#include<algorithm>