冒泡排序

时间复杂度：O（n^2）.

#include<iostream>
using namespace std;
int main()
{
    int a[5]={5,4,3,2,1};
    int dian=5;
    while(dian>1)
    {
        for(int i=0;i<dian-1;i++)
        {
            if(a[i]>a[i+1])
            {
                int tmp=a[i];
                a[i]=a[i+1];
                a[i+1]=tmp;
            }
        }
        dian--;
    }
    for(int i=0;i<5;i++)
        cout<<a[i]<<endl;
    return 0;
}