思路:1、排序,取前k个元素;O(NlogN);2、分治,O(n),利用快排的思想;3、用set 维护最小的k个数,O(NlogK),可处理海量数据。
#include <iostream> using namespace std; void print(int *a,int n){ if(a==NULL || n<=0 ) return; for(int i=0;i<n;i++){ cout<<a[i]<<" "; } cout<<endl; } int Partition(int *a,int left,int right){ int p=a[left]; int l=left; int r=right; while(l<r){ while( l<r && a[r]>=p) r--; if(l<r){ a[l]=a[r]; l++; } while( l<r && a[l]<=p) l++; if(l<r){ a[r]=a[l]; r--; } } a[l]=p; return l; } void quick_sort(int *a,int left,int right){ if(left>=right) return; int p=Partition(a,left,right); quick_sort(a,left,p-1); quick_sort(a,p+1,right); } void leastKnum(int *a,int left,int right,int k){ if(left>=right) return; int l=left; int r=right; int p=Partition(a,left,right); while(p!=k-1){ if(p>k-1){ r=p-1; p=Partition(a,l,r); } else{ l=p+1; p=Partition(a,l,r); } } for(int i=0;i<k;i++) { cout<<a[i]<<" "; } cout<<endl; } int main(){ int a[10]={2,7,8,3,5,4,9,0,1,6}; print(a,10); leastKnum(a,0,9,4); //quick_sort(a,0,5); //print(a,6); return 0; }
原文:http://blog.csdn.net/dutsoft/article/details/26704699