LeetCode专题-模拟

栏目: 编程工具 · 发布时间: 6年前

内容简介:EasyOn an infinite plane, a robot initially stands at (0, 0) and faces north. The robot can receive one of three instructions:The robot performs the instructions given in order, and repeats them forever.

1041. Robot Bounded In Circle

Easy

On an infinite plane, a robot initially stands at (0, 0) and faces north. The robot can receive one of three instructions:

"G": go straight 1 unit;
"L": turn 90 degrees to the left;
"R": turn 90 degress to the right.

The robot performs the instructions given in order, and repeats them forever.

Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

Example 1:

Input: "GGLLGG"

Output: true

Explanation:

The robot moves from (0,0) to (0,2), turns 180 degrees, and then returns to (0,0).

When repeating these instructions, the robot remains in the circle of radius 2 centered at the origin.

题目大意:对一个机器人发送一系列重复的指令,要求判断机器人会不会回到原点。

解题思路:指令是重复的,因此只要机器人不朝向北方,就有机会回到原点

class Solution:
    def isRobotBounded(self, instructions: str) -> bool:
        x = 0
        y = 0
        dir = 0 #direction - N W S E
        #offset for each direction
        dx = [0, -1, 0, 1]
        dy = [1, 0, -1, 0]
        for c in instructions:
            if c == 'G': #go ahead by dir
                x += dx[dir]
                y += dy[dir]
            elif c == 'L': #turn left
                dir = (dir + 3)%4
            elif c == 'R': #turn right
                dir = (dir + 1)%4
        return (x == 0 and y == 0) or dir != 0 #if go back to origin or not facing north

测试一下

Success
[Details]
Runtime: 32 ms, faster than 86.36% of Python3 online submissions for Robot Bounded In Circle.
Memory Usage: 13.2 MB, less than 100.00% of Python3 online submissions for Robot Bounded In Circle.

838. Push Dominoes

Medium

There are N dominoes in a line, and we place each domino vertically upright.

In the beginning, we simultaneously push some of the dominoes either to the left or to the right.

LeetCode专题-模拟

image

After each second, each domino that is falling to the left pushes the adjacent domino on the left.

Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right.

When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces.

For the purposes of this question, we will consider that a falling domino expends no additional force to a falling or already fallen domino.

Given a string "S" representing the initial state. S[i] = 'L' , if the i-th domino has been pushed to the left; S[i] = 'R' , if the i-th domino has been pushed to the right; S[i] = '.' , if the i -th domino has not been pushed.

Return a string representing the final state.

Example 1:

Input:".L.R...LR..L.."

Output:"LL.RR.LLRRLL.."

Example 2:

Input:"RR.L"

Output:"RR.L"

Explanation:The first domino expends no additional force on the second domino.

题目大意:对于一组多米诺骨牌,给定一个初始化推的指令,求出最终多米诺骨牌的状态。

解题思路:通过推的指令,计算每一个骨牌的状态。

class Solution(object):
    def pushDominoes(self, D):
        """
        :type dominoes: str
        :rtype: str
        """
        max_int = 9999999
        #record the distance from current dominoe to nearest push down one(left and right)
        left_steps = [max_int for x in range(len(D))]
        right_steps = [max_int for x in range(len(D))]

        for i in range(len(D)):
            if D[i] == 'L':
                #if occur one push down domine
                left_steps[i] = 0
                #for all following domine that influenced
                for j in range(i-1, -1, -1):
                    if D[j] != '.': #only influence '.'
                        break
                    left_steps[j] = left_steps[j+1]+1
            elif D[i] == 'R':
                right_steps[i] = 0
                for j in range(i+1, len(D)):
                    if D[j] != '.':
                        break
                    right_steps[j] = right_steps[j-1]+1

        #simulate the push work
        ND = ''
        for i in range(len(D)):
            if left_steps[i] < right_steps[i]:
                ND += 'L'
            elif left_steps[i] > right_steps[i]:
                ND += 'R'
            else:
                ND += '.'

        return ND

测试一下,算法应该是对的,但是用 python 实现会超时,用golang或者cpp可以通过所有测试用例。

Time Limit Exceeded
[Details]

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持 码农网

查看所有标签

猜你喜欢:

本站部分资源来源于网络,本站转载出于传递更多信息之目的,版权归原作者或者来源机构所有,如转载稿涉及版权问题,请联系我们

百面机器学习

百面机器学习

诸葛越、葫芦娃 / 人民邮电出版社 / 2018-8-1 / 89.00元

人工智能领域正在以超乎人们想象的速度发展,本书赶在人工智能彻底占领世界之前完成编写,实属万幸。 书中收录了超过100道机器学习算法工程师的面试题目和解答,其中大部分源于Hulu算法研究岗位的真实场景。本书从日常工作、生活中各种有趣的现象出发,不仅囊括了机器学习的基本知识 ,而且还包含了成为出众算法工程师的相关技能,更重要的是凝聚了笔者对人工智能领域的一颗热忱之心,旨在培养读者发现问题、解决问......一起来看看 《百面机器学习》 这本书的介绍吧!

图片转BASE64编码
图片转BASE64编码

在线图片转Base64编码工具

MD5 加密
MD5 加密

MD5 加密工具

HEX HSV 转换工具
HEX HSV 转换工具

HEX HSV 互换工具