Sorting is often useful as the first step in many different tasks. The most common task is to make finding things easier, but there are other uses also.
Challenge
Given a list of unsorted numbers, can you find the numbers that have the smallest absolute difference between them? If there are multiple pairs, find them all.
Input Format
There will be two lines of input:
- n - the size of the list
- array - the n numbers of the list
Output Format
Output the pairs of numbers with the smallest difference. If there are multiple pairs, output all of them in ascending order, all on the same line (consecutively) with just a single space between each pair of numbers. If there‘s a number which lies in two pair, print it two times (see sample case #3 for explanation).
Constraints
10 <= n <= 200000
-(107) <= x <= (107), where x ∈ array
array[i] != array[j], 0 <= i, j < N, and i != j
水题：维护一个ArrayList和记录当前最小距离的变量miniDist，每当i和i+1两个数的距离和miniDist相等时，把i放到ArrayList里面；当两个数距离小于miniDist时，把ArrayList清空，i放进去，并更新miniDist；最后根据ArrayList输出答案。
代码如下：
1 import java.util.*;
2
3 public class Solution {
4 public static void main(String[] args) {
5 Scanner in = new Scanner(System.in);
6 int n = in.nextInt();
7 int[] ar = new int[n];
8 for(int i = 0;i < n;i++)
9 ar[i]= in.nextInt();
10
11 Arrays.sort(ar);
12 ArrayList<Integer> index = new ArrayList<Integer>();
13 int miniDis = Integer.MAX_VALUE;
14 for(int i = 0;i < n-1;i++){
15 if(ar[i+1]-ar[i] < miniDis){
16 index.clear();
17 index.add(i);
18 miniDis = ar[i+1]-ar[i];
19 }
20 else if(ar[i+1]-ar[i]==miniDis)
21 index.add(i);
22 }
23 for(int i:index){
24 System.out.printf("%d %d ", ar[i],ar[i+1]);
25 }
26 System.out.println();
27
28 }
29 }
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