单调栈 O(n) AC
思路：
考虑暴力统计的话，时间复杂度是O(n^2)
故考虑使用单调栈，当需要寻找第i个元素之前的首个小于a[i]的元素，维护一个单调上升的栈

import java.util.Scanner;

public class Main {
    int N = 100000 + 10;
    int[] st = new int[N];
    int tt = -1;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] nums = new int[n];

        for (int i = 0; i < n; i++) nums[i] = sc.nextInt();

        Main ins = new Main();
        int[] ans = ins.solve(nums);

        for (int i = 0; i < n; i++) System.out.print(ans[i] + " ");
        sc.close();
    }

    private int[] solve(int[] a) {
        int n = a.length;
        int[] ans = new int[n];

        for (int i = 0; i < n; i++) {
            while (tt != -1 && st[tt] >= a[i]) tt--;
            if (tt == -1) {
                ans[i] = -1;
            } else {
                ans[i] = st[tt];
            }
            tt++;
            st[tt] = a[i];
        }

        return ans;
    }
}