【一次过】Lintcode 1104. Judge Route Circle

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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 (Up) and D (down). The output should be true or false representing whether the robot makes a circle.

样例

Example 1:

Input: "UD"
Output: true

Example 2:

Input: "LL"
Output: false

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解题思路：

大致的题意是：不管怎么动，判断最终是否回到了原点。由于只有四个方向，两两为一组，即Up,Down一组，Left,Right一组，同一组方向上有影响，不同组方向上不影响。

所以使用两个栈来存放UD和LR，以UD这组为例，当U压入栈时，栈为空或者栈顶元素为U，则可以将U压入，栈顶为D，则可以抵消，将栈顶弹出。遍历到最后，如果两个栈仍然为空，则表明两组方向上两两抵消，最终肯定会回到原点。

所以这题看着很多，其实与括号匹配的题目非常类似。

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<>(); //用于存储UD
        Stack<Character> stk2 = new Stack<>(); //用于存储LR
        
        for(int i=0 ; i<moves.length() ; i++){
            if(moves.charAt(i) == 'U' || moves.charAt(i) == 'D'){ //UD情况
                if(stk1.empty() || stk1.peek()==moves.charAt(i)){ 
                    stk1.push(moves.charAt(i));
                }else{ //stk1栈顶为相反的情况
                    stk1.pop();
                }
            }else{ //LR情况
                if(stk2.empty() || stk2.peek()==moves.charAt(i)){ 
                    stk2.push(moves.charAt(i));
                }else{ //stk2栈顶为相反的情况
                    stk2.pop();
                }
            }
        }
        
        if(stk1.empty() && stk2.empty())
            return true;
        else
            return false;
    }
}