A two-dimensional array of maximum connectivity sub-arrays and

1. Design idea: The user keyboard input a custom several rows of a few columns of the matrix, and then enter the relevant values, and then call the two-dimensional array of maximum connectivity sub-array and the method, the final output maximum value.

2. There is a problem: the idea of a sub-array of the unicom is not very understanding, unclear.

3. Solution: The maximum and minimum blocks of each row are first asked, then the total maximum value is then output; when the number of columns is 1 o'clock, it is calculated as one dimension;

4. Source code:

/*returns the and of the largest sub-array in an integer array. Enter a one-dimensional shape array with positive and negative numbers in the array. One-dimensional arrays end-to-end, like one end-to-end tape. One or more consecutive integers in an array make up a sub-array, each of which has a and. The maximum value for the and of all sub-arrays. */ PackageShuzumax;ImportJava.util.Scanner; Public classshuzuu{ Public Static voidMain (string[] args) {Scanner Scanner=NewScanner (system.in); System.out.println ("Please enter the rows and columns of the two-dimensional array:"); inthang=Scanner.nextint (); intLie=Scanner.nextint (); int[] shuzu=New int[Hang][lie]; inti,j;  for(i=0;i)        {             for(j=0;j<lie;j++) {Shuzu[i][j]=Scanner.nextint (); }} System.out.println ("The maximum number of consecutive and continuous values in this array is:" +Max (Shuzu,hang,lie)); }      Static intMaxintShuzu[][],intHangintlie) {        intMax=0,sum=0; inti,j; int[] b=New int[Lie]; int[] up=New int[Hang]; int[] down=New int[Hang]; int[] t=New int[Hang]; if(lie==1)        {             for(i=0;i) {sum+=shuzu[i][0]; if(sum<0) {sum=0; }                if(sum>max) {Max=sum; }            }           if(max==0)           {                for(i=0;i)               {                   if(i==0) {Max=shuzu[i][0]; }                   if(shuzu[i][0]>max) {Max=shuzu[i][0]; }               }            }        }        Else        {             for(i=0;i)            {                 for(j=0;j<lie;j++) {B[j]=Shuzu[i][j]; }                int[] c=New int[100]; c[0]=0; intSum1=0,max1=0, K;  for(k=0;k<lie;k++)                {                    if(sum1<0) {sum1=B[k]; }                    Else{sum1=sum1+B[k]; } C[k]=sum1; } max1=c[0]; intMmark=0,smark=0;  for(k=0;k<lie;k++)                {                    if(max1<C[k]) {Max1=C[k]; Mmark=K; }                }                 for(k = mmark;k >= 0;k--)                {                    if(C[k] = =B[k]) {Smark=K;  Break; }} sum=Max1; Up[i]=Smark; Down[i]=Mmark; T[i]=sum; }            intT2=t[0];  for(i=0;i)            {                if(Up[i]<=down[i+1] && down[i]>=up[i+1]) {T2+=t[i+1]; }                 for(j=up[i];j<up[i+1];j++)                {                    if(shuzu[i+1][j]>0) t2+=shuzu[i+1][j];//distinguishing independent positive numbers}} Max=T2; }        returnMax; }} 

5.:

6. Summary: Through this experiment, I think I understand the two-dimensional array is not deep enough, for the problem to a layer of analysis, the problem can be simplified, in order to progress.

A two-dimensional array of maximum connectivity sub-arrays and