leetcode 657. Robot Return to Origin（水）

最新推荐文章于 2020-01-03 13:56:58 发布

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本文探讨了机器人在二维平面上，根据一系列指令移动后能否返回原点的问题。通过解析指令字符串，利用水平和垂直位移计数，判断最终位置是否为(0,0)。

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.

题意：

机器人可以上下左右移动，起始点（0，0），执行一串指令后是否能回到原点

知识点：

再次记忆：for(char move : moves.toCharArray()){

bug:

在最后判定条件总想省事不写两个都为0

class Solution {
    public boolean judgeCircle(String moves) {
        int level = 0;
        int vertical = 0;
        for(char move : moves.toCharArray()){
            if(move == 'U'){
                vertical++;
            }else if(move == 'D'){
                vertical--;
            }else if(move == 'L'){
                level--;
            }else{
                level++;
            }
        }
        if((vertical == 0) && (level == 0)){
            return true;
        }else{
            return false;
        }
    }
}