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：
