题解 | #最大间隔#

暴力解法：记录删除掉a1 ... an-1后的每一个数组的最大间隔，取最小值

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            int n = in.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
            }
            //保存所有情况下的最大间隔
            List<Integer> ans = new ArrayList<>();
            for (int i = 1; i < n - 1; i++) {
                //获取删除掉当前元素后的数组
                List<Integer> curList = removeList(arr, i);
                int maxDiff = 0;//记录此时数组的最大间隔
                for (int j = 1; j < curList.size(); j++) {
                    int diff = curList.get(j) - curList.get(j - 1);
                    maxDiff = Math.max(maxDiff, diff);
                }
                ans.add(maxDiff);
            }
            //此时ans中记录了所有情况下的最大间隔，找到最小值即可
            Collections.sort(ans);
            System.out.println(ans.get(0));
        }
    }
    //获取删除掉某一个元素后的数组
    public static List<Integer> removeList(int[] arr, int index) {
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) {
            if (i == index) continue;
            res.add(arr[i]);
        }
        return res;
    }
}