Minimum number of k C + + (bfprt, heap sort)

Title Description Enter n integers to find the smallest number of K. For example, enter 4,5,1,6,2,7,3,8 these 8 numbers, then the smallest 4 number is 1,2,3,4,.

//Heap Sort#include <iostream>#include<algorithm>#include<vector>using namespacestd;classsolution{ Public:    voidHeapadjust (vector<int> &a,intSintN) {intRc=A[s];  for(intj=2*s+1; j<=n-1; j=j*2+1)        {            if(j<n-1&&a[j]>a[j+1]) J++; if(rc<=A[j]) Break; Else{A[s]=A[j]; S=J; }} A[s]=RC; }    voidCreatheap (vector<int> &a,intN) { for(inti=n/2-1; i>=0; i--) {heapadjust (a,i,n); }} vector<int> Getleastnumbers_solution (vector<int> Input,intk) {vector<int>ans; if(input.size () = =0|| k==0|| K>input.size ())returnans; intn=input.size ();        Creatheap (Input,n);  for(inti=n-1; I>0; i--)        {                intx=input[0]; input[0]=Input[i]; Input[i]=x; Heapadjust (Input,0, i); }         for(intI=input.size ()-1; i>=0&&k>0; i--) {ans.push_back (input[i]); K--; }        returnans; }};intMain () {solution S; intk=4; Vector<int>ans; Vector<int>input; Input.push_back (4); Input.push_back (5); Input.push_back (1); Input.push_back (6); Input.push_back (2); Input.push_back (7); Input.push_back (3); Input.push_back (8); Ans=s.getleastnumbers_solution (input,k);  for(intI=0; I<ans.size (); i++) {cout<<Ans[i]; }    return 0;}

//bfprt#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>using namespacestd;classSolution { Public: Vector<int> Getleastnumbers_solution (vector<int> Input,intk) {vector<int>ans; if(input.size () = =0|| k==0|| K>input.size ()) {            returnans; }         for(intI=0; I&LT;=BFPRT (Input,0, Input.size ()-1, k); i++) {ans.push_back (input[i]); }        returnans; }    /*Inserts a sort, returns the middle digit subscript*/    intInsertsort (vector<int> &array,intLeftintRight ) {        inttemp; intJ;  for(inti = left +1; I <= right; i++) {Temp=Array[i]; J= i-1;  while(j >= left && Array[j] >temp) Array[j+1] = array[j--]; Array[j+1] =temp; }        return((Right-left) >>1) +Left ; }    /*returns The median subscript for the median number of digits*/    intGetpivotindex (vector<int> &array,intLeftintRight ) {        if(Right-left <5)            returninsertsort (array, left, right); intSub_right = left-1;  for(inti = left; i +4<= right; i + =5)        {            intindex = insertsort (array, I, i +4);//Find the subscript for the median of five elementsSwap (Array[++sub_right], array[index]);//On the left, in turn        }        returnBfprt (array, left, Sub_right, (Sub_right-left +1) >>1) +1); }    /*Using the median median of the subscript to divide, return to the dividing line subscript*/    intPartition (vector<int> &array,intLeftintRightintPivot_index)  {Swap (Array[pivot_index], array[right]); //Place the datum at the end        intDivide_index = left;//the dividing line of tracking Division         for(inti = left; I < right; i++)        {            if(Array[i] <Array[right]) swap (Array[divide_index+ +], array[i]);//smaller than the benchmark, on the left.} swap (Array[divide_index], array[right]); //Finally, return the benchmark .        returnDivide_index; }    intBFPRT (vector<int> &array,intLeftintRightConst int&k) {intPivot_index = Getpivotindex (array, left, right);//to get The median of the middle-digit subscript        intDivide_index = Partition (array, left, right, Pivot_index);//to divide and return to a dividing boundary        intnum = Divide_index-left +1; if(num = =k)returnDivide_index; Else if(Num >k)returnBfprt (array, left, Divide_index-1, K); Else            returnBfprt (Array, Divide_index +1, right, K-num); }};intMain () {solution S; intk=4; Vector<int>ans; Vector<int>input; Input.push_back (4); Input.push_back (5); Input.push_back (1); Input.push_back (6); Input.push_back (2); Input.push_back (7); Input.push_back (3); Input.push_back (8); Ans=s.getleastnumbers_solution (input,k);  for(intI=0; I<ans.size (); i++) {cout<<Ans[i]; }    return 0;}

Minimum number of k C + + (bfprt, heap sort)