题目描述

朴素版上升子序列

import java.util.*;

public class Main {
    static final int N = 1010;
    static int[] f = new int[N], a = new int[N];
    static int n;

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

        n = sc.nextInt();

        for (int i = 1; i <= n; i++) a[i] = sc.nextInt();

        for (int i = 1; i <= n; i++) {
            f[i] = 1;
            for (int j = i - 1; j >= 1; j--) {
                if (a[j] < a[i]) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
        }

        int res = 0;
        for (int i = 1; i <= n; i++) 
            if (res < f[i]) res = f[i];

        System.out.println(res);
    }
}

贪心

$O(n^logn)$

下标从0开始

import java.util.*;

public class Main {
    static final int N = 100010;
    static int[] arr = new int[N], q = new int[N];
    static int n;

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        n = sc.nextInt();
        sc.nextLine();

        String[] s = sc.nextLine().split(" ");
        for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(s[i]);

        int len = 0;
        q[0] = (int) -2e9;
        for (int i = 0; i < n; i++) {
            int l = 0, r = len;
            while (l < r) {
                int mid = l + r + 1 >> 1;
                if (q[mid] < arr[i]) l = mid;
                else r = mid - 1;
            }
            len = Math.max(len, l + 1);
            q[l + 1] = arr[i];
        }

        System.out.println(len);
    }
}

下标从1开始

import java.util.*;

public class Main {
    static final int N = 100010;
    static int[] arr = new int[N], q = new int[N];
    static int n;

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        n = sc.nextInt();
        sc.nextLine();

        String[] s = sc.nextLine().split(" ");
        for (int i = 1; i <= n; i++) arr[i] = Integer.parseInt(s[i - 1]);

        int len = 0;
        //q[i]表示长度为i的上升子序列的最小头结点：
        //对于1 2 3 4：q[3]:长度为3的上升子序列的最小值即为3也就是q[i]=3
        for (int i = 1; i <= n; i++) {
            //寻找a[i]要插入的位置
            int l = 0, r = len;
            while (l < r) {
                int mid = l + r + 1 >> 1;
                if (q[mid] < arr[i]) l = mid;
                else r = mid - 1;
            }
            len = Math.max(len, l + 1);
            q[l + 1] = arr[i];
        }

        System.out.println(len);
    }
}