bubble sort algorithm

• simple but inefficient
• Scan the data from left to right, select the largest data, put it on the right
• Key point: Compare two adjacent numbers and swap if the number on the left is greater than the number on the right
• Assuming that there are n elements waiting to be compared in the array, when the first sorting is over, the largest element will be moved to the end of the sequence; when the second sorting is over, the second largest element will be moved to the n- of the sequence. 1 position; after n-1 sorting, the bubble sorting is completed~

The C implementation is as follows: requires 2 loops

One trip: 1 with 2, 2 with 3, 3 with 4, 4 with 5, 5 with 6, 6 with 7, 7 with 8, 8 with 9, 9 with 10: the back is higher than the front, and the bottom is lower than the front. No transposition - swap the largest to the right for a total of 9 comparisons

2 times: 1 with 2, 2 with 3, 3 with 4, 4 with 5, 5 with 6, 6 with 7, 7 with 8, 8 with 9: the back is higher than the front, and the bottom is lower than the front, no transposition— ——Change the next largest to n-1 bit for a total of 8 comparisons

So ~ every time the outer loop is completed, the number of comparisons in the inner loop is reduced by one

#include<stdlib.h>
#include<stdio.h>

void bubblesort(int *a, int n);

intmain()
{
	int k;
	int a[10] = {2,4,6,8,0,1,3,5,7,9};
	bubblesort(a, 10);
	for (k = 0; k < 10; k++)
		printf("%d\t", a[k]);
	system("pause");
	return 0;
}

void bubblesort(int *a,int n)
{
	int i, j, t;
	//10 color bars, 9 rounds of comparison, the outer loop controls the number of rounds of comparison
	for (i = 0; i < n - 1; i++)
	{
		//Every time the outer loop is performed once, the comparison of the inner loop is reduced once, so j<ni-1
		for (j = 0; j < n - i - 1; j++)
		{
			if (a[j]>a[j + 1])
			{
				t = a[j ];
				a[j] = a[j + 1];
				a[j + 1] = t;
			}
		}
	}
}