Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place finally.

The move sequence is represented by a string. And each move is represent by a character. The valid robot moves are R (Right), L (Left), U

樣例

Example 1:

Input: "UD"
Output: true

Example 2:

Input: "LL"
Output: false

解題思路：

大致的題意是：不管怎麼動，判斷最終是否回到了原點。由於只有四個方向，兩兩為一組，即Up,Down一組，Left,Right一組，同一組方向上有影響，不同組方向上不影響。

所以使用兩個棧來存放UD和LR，以UD這組為例，當U壓入棧時，棧為空或者棧頂元素為U，則可以將U壓入，棧頂為D，則可以抵消，將棧頂彈出。遍歷到最後，如果兩個棧仍然為空，則表明兩組方向上兩兩抵消，最終肯定會回到原點。

所以這題看著很多，其實與括號匹配的題目非常類似。

public class Solution {
    /**
     * @param moves: a sequence of its moves
     * @return: if this robot makes a circle
     */
    public boolean judgeCircle(String moves) {
        // Write your code here
        if(moves.length()==0)
            return true;
        
        Stack<Character> stk1 = new Stack<>(); //用於儲存UD
        Stack<Character> stk2 = new Stack<>(); //用於儲存LR
        
        for(int i=0 ; i<moves.length() ; i++){
            if(moves.charAt(i) == 'U' || moves.charAt(i) == 'D'){ //UD情況
                if(stk1.empty() || stk1.peek()==moves.charAt(i)){ 
                    stk1.push(moves.charAt(i));
                }else{ //stk1棧頂為相反的情況
                    stk1.pop();
                }
            }else{ //LR情況
                if(stk2.empty() || stk2.peek()==moves.charAt(i)){ 
                    stk2.push(moves.charAt(i));
                }else{ //stk2棧頂為相反的情況
                    stk2.pop();
                }
            }
        }
        
        if(stk1.empty() && stk2.empty())
            return true;
        else
            return false;
    }
}