zoukankan      html  css  js  c++  java
  • [LeetCode] 657. Robot Return to Origin

    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: moves = "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: moves = "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.
    

    Example 3:

    Input: moves = "RRDD"
    Output: false
    

    Example 4:

    Input: moves = "LDRRLRUULR"
    Output: false

    Constraints:

    • 1 <= moves.length <= 2 * 104
    • moves only contains the characters 'U''D''L' and 'R'.

    机器人能否返回原点。

    这道题不涉及算法,思路就是模拟,直接上代码了。

    时间O(n)

    空间O(n) - charArray

    Java实现

     1 class Solution {
     2     public boolean judgeCircle(String moves) {
     3         char[] steps = moves.toCharArray();
     4         int up = 0;
     5         int left = 0;
     6         for (char step : steps) {
     7             if (step == 'U') {
     8                 up++;
     9             } else if (step == 'D') {
    10                 up--;
    11             } else if (step == 'L') {
    12                 left++;
    13             } else if (step == 'R') {
    14                 left--;
    15             }
    16         }
    17         return up == 0 && left == 0;
    18     }
    19 }

    LeetCode 题目总结

  • 相关阅读:
    CF1328B K-th Beautiful String
    CF1327B Princesses and Princes
    CF750D New Year and Fireworks
    CF57C Array
    洛谷P5661 公交换乘(CSP-J 2019 T2)
    Docker原理:Cgroup
    Docker原理:Namespace
    Anaconda软件安装使用问题
    初步了解Unix系统的I/O模式
    深入理解索引和AVL树、B-树、B+树的关系
  • 原文地址:https://www.cnblogs.com/cnoodle/p/13575052.html
Copyright © 2011-2022 走看看