LeetCode Maximum Distance in Arrays

原题链接在这里：https://leetcode.com/problems/maximum-distance-in-arrays/description/

题目：

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].

题解：

从第一行的array 拿出首尾两值作为min, max value. 接下来的array 中，最大的distance只能从max-array.get(0) 和 array.get(array.size()-1) - min中选较大值, 同时更新min 和 max值. 维护最大的distance在res中.

Time Complexity: O(n). n = arrays.size().

Space: O(1).

AC Java:

 1 public class Solution {
 2     public int maxDistance(List<List<Integer>> arrays) {
 3         if(arrays == null || arrays.size() <= 1){
 4             return 0;
 5         }
 6         
 7         int res = 0;
 8         int min = arrays.get(0).get(0);
 9         int max = arrays.get(0).get(arrays.get(0).size()-1);
10         for(int i = 1; i<arrays.size(); i++){
11             res = Math.max(res, Math.max(max - arrays.get(i).get(0), arrays.get(i).get(arrays.get(i).size()-1)-min));
12             max = Math.max(max, arrays.get(i).get(arrays.get(i).size()-1));
13             min = Math.min(min, arrays.get(i).get(0));
14         }
15         return res;
16     }
17 }

 

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/7362374.html