LeetCode 657.Robot Return to Origin-优快云博客

本文链接：https://blog.youkuaiyun.com/geekdoe/article/details/104980287

本文介绍了一种算法，用于判断二维平面上的机器人在完成一系列移动指令后是否能返回到起点(0,0)。通过分析移动序列，算法检查向左和向右、向上和向下的移动次数是否相等，以此来确定机器人是否能回到原点。

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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.

//java
class Solution {
    public boolean judgeCircle(String moves) {
        //如果长度为奇数，则不可能回到原点
        if((moves.length() >> 1) << 1 != moves.length() )
            return false;
        char []ch = moves.toCharArray();
        int []arr = new int[126];
        for(char st : ch){
            arr[st]++;
        }
        if(arr['U'] == arr['D'] && arr['L'] == arr['R'])
            return true;
        else
            return false;
    }
}