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; }