Ordering of Java Data structures

1. Bubble Sort: Time complexity is O (n2)

The hypothesis is a small to large sort: a comparison between two adjacent numbers, and a larger number at the back. The largest number in the last row after a comparison

such as: 40, 8, 15, 18, 121 times after the order is: 8, 15, 18, 12, 40 in turn until from small to big pat good

 for (int i = 0;i < n-i; i++) {   for (int j = 0;j <n-1-i;j++) {//The last few are already lined up There is no need to sort       the if(A[j] > a[j+1]) {           k= a[j]           ; = A[j+1];           A[j+1] = k;}}  }

2. Select Sort: Time complexity O (n2)

The hypothesis is a sort of small to large: each time the smallest number in a sort order is placed first

such as: 3, 4, 1, 5, 21 times after sorting: 1, 4, 3, 5, 2

 for (int i=0;i<n-1;i++) {    for (int j=i+1;j<n;j++) {        if (a[i]>a[j]) {           k= a[i];            = A[j];            = k;}}    }    

3. Insert Sort:

The assumption is a small-to-large sort: each time a number is compared with the previous number, if it is less than the preceding number

 for (int I =1;i < n; i++) {    int j = i-1;     int k = a[i];      while (A[i] < a[j]&&j >= 0) {         a[j+1] = A[j];
         j--;    }     = k;}

Ordering of Java Data structures