zoukankan      html  css  js  c++  java
  • interviewbit :Min Steps in Infinite GridBookmark Suggest Edit

    You are in an infinite 2D grid where you can move in any of the 8 directions :

       (x,y) to 
        (x+1, y), 
        (x - 1, y), 
        (x, y+1), 
        (x, y-1), 
        (x-1, y-1), 
        (x+1,y+1), 
        (x-1,y+1), 
        (x+1,y-1) 
    

    You are given a sequence of points and the order in which you need to cover the points. Give the minimum number of steps in which you can achieve it. You start from the first point.

    Example :

    Input : [(0, 0), (1, 1), (1, 2)]
    Output : 2
    

    It takes 1 step to move from (0, 0) to (1, 1). It takes one more step to move from (1, 1) to (1, 2).

    This question is intentionally left slightly vague. Clarify the question by trying out a few cases in the “See Expected Output” section.

    public class Solution {
        // X and Y co-ordinates of the points in order.
        // Each point is represented by (X.get(i), Y.get(i))
        public int coverPoints(ArrayList<Integer> X, ArrayList<Integer> Y) {
            int x=0;
            int i,j,k;
            for(i=1;i<X.size();++i){
                j=Math.abs(X.get(i)-X.get(i-1));
                k=Math.abs(Y.get(i)-Y.get(i-1));
                x+=(j>k)?j:k;
            }
            return x;
    
        }
    }
  • 相关阅读:
    Java回调机制
    显示Title和隐藏Title的ListView
    ListView的小知识
    小知识点
    Shader的使用
    项目知识(二)
    项目知识(一)
    复习篇(一)Activity的生命周期和启动模式
    BootStrap 提示框
    BootStrap选项卡
  • 原文地址:https://www.cnblogs.com/bbbblog/p/5724810.html
Copyright © 2011-2022 走看看