10. Robot Return to Origin

Title：

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.

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:

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.

Analysis of Title：

It's easy. But there are some interesting methods.

Test case：

"UD"

Python：

Method one: Go through 4 times.

　　return moves.count('L') == moves.count('R') and moves.count('U') == moves.count('D')  

Mehtod two: Gothrough 1 times.

u,d,r,l=1,1,1,1
for i in moves:
　　if i=='U':
　　　　u+=1
　　elif i=='D':
　　　　d+=1
　　elif i=='R':
　　　　r+=1
　　else:
　　　　l+=1
　　if d==u and r==l:
　　　　return True
　　return False

But this method uses too many variables and below this is better.

x, y = 0, 0
for i in moves:
　　if i == "U":
　　　　y += 1
　　elif i == "D":
　　　　y -= 1
　　elif i == "R":
　　　　x += 1
　　elif i == "L":
　　　　x -= 1
　　if x == 0 and y == 0:
　　　　return True
　　else:
　　　　return False

Analysis of Code：

Use less variables.