题解 | #信封嵌套#

根据题意，将第一维的数据进行排序，第二维的数据防止重复计算进行逆序排序，转化为求最长的上升子序列的长度(O(nlogn))的时间复杂度。

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        int n = reader.nextInt();
        int[][] letters = new int[n][2];
        int[] tmp = new int[n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 2; j++) {
                letters[i][j] = reader.nextInt();
            }
        }
        // 贪心 按照第一个进行排序，按照第二列进行最长上升序列
        Arrays.sort(letters, (n1, n2) -> {
            if (n1[0] != n2[0]) {
                return n1[0] - n2[0];
            } else {
                return n2[1] - n1[1];
            }
        });
        
        for (int i = 0; i < n; i++) {
            tmp[i] = letters[i][1];
        }
        System.out.println(helper(tmp));
    }

    public static int helper(int[] nums) {
        int n = nums.length;
        int[] tails = new int[n + 1];
        int res = 0;
        for (int i = 0; i < n; i++) {
            int l = 0, r = res;
            while (l < r) {
                int m = (l + r) / 2;
                if (tails[m] < nums[i]) {
                    l = m + 1;
                } else {
                    r = m;
                }
            }
            tails[l] = nums[i];
            if (r == res) res += 1;
        }
        return res;
    }
}