面试30 求数组中最小的k个数

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#include<iostream>#include<cstdio>#include<cstdlib>#include<set>#include<functional>#include<algorithm>#include<ctime>using namespace std;int Partition(int *a,int start,int end){     int i=start,j=end; int key=a[i]; while(i<j) {       while(i<j&&a[j]>=key)   {        j--;   }   if(i<j)   {        a[i]=a[j];i++;   }   while(i<j&&a[i]<=key)   {        i++;   }   if(i<j)   {      a[j]=a[i];   } } a[i]=key; return i;}//以下方法时间复杂度为O（n）void GetLeastNumbers(int *input,int n,int *output,int k)//找出数组中最小的k个数{    int start=0,end=n-1;int index=Partition(input,start,end);while(index!=k-1)//跳出循环时index=k-1  此时a[0] a[1]...a[k-1] 刚好为最小的k个数{     if(index>k-1) {      end=index-1;  index=Partition(input,start,end); } else if(index<k-1) {  start=index+1;  index=Partition(input,start,end); }}cout<<"the "<<k<<"least number is:"<<endl;for(int i=0;i<k;i++){     output[i]=input[i]; cout<<output[i]<<" ";}cout<<endl;}//比较适合处理海量的数据 时间复杂度为o(logk)void GetLeastNumbers2(int *input,int n,multiset<int ,greater<int>>&LeastSet,int k)//找出数组中最小的k个数存到set中{     for(int i=0;i<n;i++) {     if(LeastSet.size()!=k) {       LeastSet.insert(input[i]); } else {     multiset<int,greater<int>>::iterator iter=LeastSet.begin(); if(*iter>input[i]) {LeastSet.erase(iter);LeastSet.insert(input[i]); } } } cout<<"the "<<k<<"least number is:"<<endl;  multiset<int,greater<int>>::iterator iter=LeastSet.begin();  for(;iter!=LeastSet.end();iter++)  {      cout<<*iter<<" ";  }  cout<<endl;}int main(){srand(time(NULL));int a[10],output[10];multiset<int,greater<int>> testSet;cout<<"the original numbers is:"<<endl;for(int i=0;i<10;i++){   a[i]=rand()%100;   cout<<a[i]<<"  ";}cout<<endl;GetLeastNumbers(a,10,output,4);GetLeastNumbers2(a,10,testSet,4);//找出数组中最小的k个数return 0;}

其实最直观的做法就是对数组从小到大排序，然后取前k个数。时间复杂度取决于排序算法

mulitset<int,greater<int>()>

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