Put all the even number in ascending order of an integer array to sort all the odd Numbers in descending order sorting, require the odd and even number
The original location does not change,


For example: array int a [5]={100,80,7,44,17}
After sorting the results as follows:
{44 reached, 17100, 7}
Sorting before even in 0, 1 and 3, even after sorting is still in the 0, 1 and 3,
Before ordering the odd in 2 and 4, after the sorting odd number is still in the 2 and 4,

Public class Csort {
Public void upbubble (int a [], int n) {
for (int i=0; i A Boolean flag=true;
for (int j=0; j If (a [j] & gt; + 1 a [j]) {
int temp;
Temp=a, [j].
A [j]=a, a + 1 bonus to [j].
A=\ [j + 1];
flag=false;
}
}
If (flag)
break;
}
}

Public void downbubble (int a [], int n) {
for (int i=0; i A Boolean flag=true;
for (int j=0; j If (a [j] <+ 1 a [j]) {
int temp;
Temp=a, [j].
A [j]=a, a + 1 bonus to [j].
A=\ [j + 1];
flag=false;
}
}
If (flag)
break;
}
}

Public void jobubble (int a [], int n) {
//finish this function
}

Public void display (int a [], int n) {
System.out.println();
for (int i=0; i System. The out. Print (a [I] + "");
System.out.println();
}

}

Public class Cmain {
Public static void main (String [] args) {
Int a, [].
A=new int [20].

A [0]=50;
A [1]=17;
A [2]=2;
A [3]=34;
A [4]=401;
A [5]=157;
A, [6]=15.
A [7]=81;
A [8]=52;
A [9]=63;
A [10]=64;
A [11]=95;
A [12]=51;
A [13]=82;
A [14]=33;
A [15]=65;
A [16]=911;
A, [17]=22.
A [18]=83;
A [19]=104;

Csort BB=new Csort ();
System. The out. Println (" original array is: ");
BB. The display (a, 20);

System. Out. Println (" odd descending even after sorting result is: ");
BB. Jobubble (a, 20);
BB. The display (a, 20);
}
}


This topic how to write ah o

CodePudding user response:

First traversal access t location information such as what you said x=,0,1,0,1 [0] 0 odd number is an even number 1 represented, and get a new array,80,44 [100] b [17, 7].
Sorting, respectively, a=b=[44,80,100] [17, 7]
Second traversal x and obtain a or b's current location data, the subscript of a and b are calculated respectively

CodePudding user response:

Thank you

Public void jobubble (int a [], int n) {
//finish this function
Int [] x=new int [n].
Int [] arr1=new int [n].
Int [] arr2=new int [n].
for(int i=0; iIf (a [I] % 2==0) {
[I] x=0;
Arr1=[I] a [I];
}
The else {
[I] x=1;
Arr2=[I] a [I];
}
}
for (int i=0; i System. The out. Print (x [I] + "");
System.out.println();
for (int i=0; i System. The out. Print (arr1 [I] + "");
System.out.println();
for (int i=0; i System. The out. Print (arr2 [I] + "");
}

Then how to separate t, in the two arrays, the result is that after I separated from
1 0 0 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 1
50 34 52 0 0 0 0 0 0 2 22 64 82 0 0 0 0 0 0, 104
17 401, 157, 15 81 0 0 0 0 0 95 63 51 0 33 65 911 0, 83

If can't achieve this function, how should do