一维数组最大子数组续（debug版）

设计思路：就是在原有一维数组中加了if条件判断一下，输入的数字是否为继续执行的数字，如果是就继续向下边执行，否则就执行指定的步骤。至于在最后输出的数据只需在循环中加入指定的数据进行记录就行。

package danbushuzu;

import java.util.Scanner;

public class danBu {
    public static void main(String[] args) throws InterruptedException {
        Scanner sc = new Scanner(System.in);
        //时间复杂度没有实现，只是实现了输出最大值
        int n=0;
        int linshin=0;
        int nshu;//循环的次数
        int shu[] = {-1,3,-6,-5,-2}; 
        int max = shu[0];//存储最大的和
        int linshimax=1;
        System.out.println("0顺序输出，其他数字为指定输出，请输入");
        for(nshu=0;nshu<5;nshu++) {
            int n1 = 0;
            int n2 = 0;
            for(int nnshu=nshu;nnshu<5;nnshu++) {
                int type = sc.nextInt();
                if(type==0) {
                    n1 = n1 + shu[nnshu];
                    int zongmax=0;
                    if(nnshu<4) {
                        zongmax=max;
                        nnshu = nnshu + 1;
                        n2 = n1 + shu[nnshu];
                        max = maxxx(n1,n2,max);
                        nnshu = nnshu - 1;
                    }else {
                        zongmax=max;
                        max = maxx(n1,max);
                    }
                    linshin++;
                    System.out.println("当前计算到第" + linshin + "个子数组");
                    if(zongmax!=max) {
                        System.out.println("最大子数组为第" + linshin + "个子数组");
                        linshimax=linshin;
                    }else {
                        System.out.println("最大子数组为第" + linshimax + "个子数组");
                    }
                    System.out.println("当前运行到第" + (nshu+1) + "个到第" + (nnshu+1) + "个");
                    System.out.println("最大值为：" + max);
                    Thread.sleep(1000);
                }else {
                    linshin=0;
                    n=type;
                    for(nnshu=nshu;nnshu<5;nnshu++) {
                        n1 = n1 + shu[nnshu];
                        int zongmax=0;
                        
                        
                        if(nnshu<4) {
                            zongmax=max;
                            nnshu = nnshu + 1;
                            n2 = n1 + shu[nnshu];
                            max = maxxx(n1,n2,max);
                            nnshu = nnshu - 1;
                        }else {
                            zongmax=max;
                            max = maxx(n1,max);
                        }
                        linshin++;
                        if(linshin==n) {
                            System.out.println("当前计算到第" + linshin + "个子数组");
                            if(zongmax==max) {
                                System.out.println("最大子数组为第" + linshimax + "个子数组");
                            }else {
                                System.out.println("最大子数组为第" + linshin + "个子数组");
                            }
                            System.out.println("当前运行到第" + (nshu+1) + "个到第" + (nnshu+1) + "个");
                            System.out.println("最大值为：" + max);
                            Thread.sleep(1000);
                            break;
                        }
                    }
                }
            }
        }
    }
    
    static int maxxx(int a,int b,int ab) {
        int max;
        if(a<b) {
            max = b;
            if (max<ab) {
                max = ab;
            }
        }else {
            max = a;
            if(max<ab) {
                max = ab;
            }
        }
        return max;
    }
    
    static int maxx(int a , int b){
        int max;
        if(a<b) {
            max = b;
        }else {
            max = a;
        }
        return max;
    }
    
    
}

截图如下：