O(n)时间复杂度求最小的k个数和第k小的数

//思路:使用快速排序的patition函数来进行处理 时间复杂度为O(n)#include<iostream>#include<cassert>using namespace std;int partition(int *ar, int len, int low, int high){int temp = ar[low];while(low < high){while(low < high && temp < ar[high]){high--;}if(low == high){break;}else{ar[low] = ar[high];}while(low < high && temp > ar[low]){low++;}if(low == high){break;}else{ar[high] = ar[low];}}ar[low] = temp;return low;}int findkmin(int *ar, int len, int k)//寻找第k小的数{assert( (ar!=NULL) && (k>0) && (k<len+1) );int low = 0;int high = len - 1;int index = partition(ar, len, low, high);while(index != k-1){if(index > k){index = partition(ar, len, low, index-1);}else{index = partition(ar, len, index+1, high);}}return ar[index];}void findkmin(int *ar, int len, int k, int *retar)//寻找最小的k个数{assert( (ar!=NULL) && (k>0) && (k<len+1) && (retar!=NULL));int low = 0;int high = len - 1;int index = partition(ar, len, low, high);while(index != k-1){if(index > k){index = partition(ar, len, low, index-1);}else{index = partition(ar, len, index+1, high);}}for(int i = 0; i<=index; ++i){retar[i] = ar[i];}}void show(int *ar ,int len){for(int i=0; i<len; ++i){cout<<ar[i]<<" ";}cout<<endl;}int main(){int ar[] = {7,3,5,6,4,1,2};int retar[32];int len = sizeof(ar)/sizeof(ar[0]);int len2 = sizeof(retar)/sizeof(retar[0]);cout<<findkmin(ar, len, 7)<<endl;cout<<findkmin(ar, len, 1)<<endl;cout<<findkmin(ar, len, 5)<<endl;//cout<<findkmin(ar, len, 0)<<endl;//cout<<findkmin(ar, len, 8)<<endl;findkmin(ar, len, 7, retar);show(retar, 7);findkmin(ar, len, 1, retar);show(retar, 1);findkmin(ar, len, 5, retar);show(retar, 5);return 0;}