2020022802

思路就是很简单的

遍历以每个数字开头时的所有子集合，然后比较，记下最大值

时间复杂度为O(n^2)

package hmk0;

import java.util.ArrayList;
import java.util.Scanner;

class TClass {
    public int value;

    public TClass() {
        value = 0;
    }

    public TClass(int i) {
        value = i;
    }
}

class Tmp {
    public static int func(ArrayList<TClass> ipt, int length) {
        int i = 0;
        int j = 0;
        int tmp = 0;
        int rtn = ipt.get(0).value;// 初始化防止全负
        boolean flag = true;// 标记环状回到开头
        for (i = 0; i < length; i++) {
            tmp = 0;
            flag = true;
            for (j = i; j < i || flag; j++) {
                if (j == length) {
                    if (i == 0)
                        break;
                    j = 0;
                    flag = false;
                }
                tmp = tmp + ipt.get(j).value;
                if (tmp > rtn) {
                    rtn = tmp;
                }
            }
        }
        return rtn;
    }

    public static void main(String args[]) {
        int l = 0;
        int i = 0;
        int tmp = 0;
        ArrayList<TClass> ipt = new ArrayList<TClass>();
        Scanner sipt = new Scanner(System.in);
        System.out.println("Enter The Length");
        l = sipt.nextInt();
        if (sipt.hasNextLine()) {
            sipt.nextLine();
        }
        for (i = 0; i < l; i++) {
            tmp = sipt.nextInt();
            if (sipt.hasNextLine()) {
                sipt.nextLine();
            }
            ipt.add(new TClass(tmp));
        }
        System.out.println("Output:" + func(ipt, l));
    }
}