题目描述

初始代码

import java.util.*;

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

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.nextLine());

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

        int res = 0;
        for (int i = 1; i <= n; i++) {
            f[i] = 1;
            for (int j = 1; j < i; j++)
                if (arr[i] > arr[j]) f[i] = Math.max(f[i], f[j] + 1);
        }

        for (int i = n; i > 0; i--) {
            g[i] = 1;
            for (int j = n; j > i; j--) 
                if (arr[i] > arr[j]) g[i] = Math.max(g[i], g[j] + 1);
        }

        for (int i = 1; i <= n; i++) res = Math.max(res, f[i] + g[i] - 1);

        System.out.println(res);
    }
}

简短

import java.util.*;

public class Main {
    static final int N = 1010;
    static int[] arr = new int[N];
    static int[] f = new int[N];

    public static void main(String[] arfs) {

        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.nextLine());

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

        int res = 0;
        for (int i = 1; i <= n; i++) {
            f[i] = 1;
            for (int j = 1; j < i; j++)
                if (arr[i] > arr[j]) f[i] = Math.max(f[i], f[j] + 1);
        }

        for (int i = n; i > 0; i--) {
            int t = f[i];
            f[i] = 1;
            for (int j = n; j > i; j--) 
                if (arr[i] > arr[j]) f[i] = Math.max(f[i], f[j] + 1);
            res = Math.max(res, t + f[i] - 1);
        }

        System.out.println(res);
    }
}

算法2

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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