zoukankan      html  css  js  c++  java
  • Maximum Distance in Array

    Given m arrays, and each array is sorted in ascending order. Now you can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers a and b to be their absolute difference |a-b|. Your task is to find the maximum distance.

    Example 1:

    Input: 
    [[1,2,3],
     [4,5],
     [1,2,3]]
    Output: 4
    Explanation: 
    One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.
    

    Note:

    1. Each given array will have at least 1 number. There will be at least two non-empty arrays.
    2. The total number of the integers in all the m arrays will be in the range of [2, 10000].
    3. The integers in the m arrays will be in the range of [-10000, 10000].
     1 public class Solution {
     2     public int maxDistance(List<List<Integer>> arrays) {
     3         int n = arrays.size();
     4         int[] small = new int[n];
     5         int[] large = new int[n];
     6         for (int i = 0; i < n; i++) {
     7             List<Integer> array = arrays.get(i);
     8             small[i] = array.get(0);
     9             large[i] = array.get(array.size() - 1);
    10         }
    11         
    12         HashMap<Integer, Integer> smallMap = new HashMap<>();
    13         HashMap<Integer, Integer> largeMap = new HashMap<>();
    14         for (int i = 0; i < n; i++) {
    15             smallMap.put(small[i], i);
    16             largeMap.put(large[i], i);
    17         }
    18         
    19         Arrays.sort(small);
    20         Arrays.sort(large);
    21         
    22         int low = 0, high = n - 1;
    23         if (smallMap.get(small[low]) == largeMap.get(large[high])) {
    24             
    25             // choose the second smallest
    26             int result1 = calculateABS(large[high], small[low]);
    27             if (low < n - 1) {
    28                 result1 = calculateABS(large[high], small[low + 1]);
    29             }
    30             
    31             // choose the second largest
    32             int result2 = calculateABS(large[high], small[low]);
    33             if (high > 0) {
    34                 result2 = calculateABS(large[high - 1], small[low]);
    35             }
    36             
    37             return Math.max(result1, result2);
    38         }
    39         return calculateABS(large[high], small[low]);
    40     }
    41     
    42     private int calculateABS(int num1, int num2) {
    43         return Math.abs(num1 - num2);
    44     }
    45 }
     
  • 相关阅读:
    shell中的交互模式:expect
    hive(II)--sql考查的高频问题
    ETL工具--kettle篇(17.10.09更新)
    hive(I)--学习总结之常用技能
    ubantu上搭建hive环境
    shell实例练习+详解
    搭建hadoop、hdfs环境--ubuntu(完全分布式)
    oracle 获取一个字段的年月日
    oracle 两表更新 报错ORA-01779: 无法修改与非键值保存表对应的列
    oracle 查看表空间 添加数据文件
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/7053683.html
Copyright © 2011-2022 走看看