《编程之法》：寻找最小的k个数

来源：互联网发布：南通cnc数控编程培训编辑：程序博客网时间：2024/05/16 15:18

题目

标题

有n个整数，请找出其中最小的k个数。

思路

维护一个容量为k的最大堆，用来存储最先遍历的k个数。建好堆后遍历剩余n-k个整数，将每次遍历到的新元素与堆顶做比较，若比堆顶小，则用该元素替换堆顶，然后更新堆。遍历结束后堆中的k个数即是最小的k个数。

代码

//// Created by huxijie on 17-3-18.// 寻找最小的k个数#include <iostream>#include <random>using namespace std;bool myless(int result[],int i,int j) {    if (result[i]<result[j]) return true;    else return false;}void exch(int result[],int i, int j) {    int tmp = result[i];    result[i] = result[j];    result[j] = tmp;}//由上至下的堆有序化,下沉void sink(int result[],int m,int n) {    while (2 * m + 1 <= n - 2) {        int j = 2 * m + 1;        if (j + 1 <= n-1 && myless(result, j, j + 1)) {            j = j + 1;        }        if (!myless(result, m, j)) {            break;        } else {            exch(result, m, j);            m = j;        }    }}int *FindMinK(int numbers[],int n,int k) {    int *result = new int[k];    for (int i = 0; i < k; ++i) {        result[i] = numbers[i];    }    //堆排序    for (int j = (k - 1) / 2; j >=0 ; --j) {        sink(result, j, k);    }    //遍历剩下的n-k个数,每次取出堆顶进行比较,若有更改最大值,则调整堆    for (int i = k; i< n; ++i) {        if (result[0] > numbers[i]) {            result[0] = numbers[i];            sink(result, 0, k);        }    }    return result;}int main() {    default_random_engine e;    e.seed(time(0));    uniform_int_distribution<int> u(-10, 100);    int n = 10;    int numbers[n];    cout<<"original numbers: ";    for (int i = 0; i < n; ++i) {        numbers[i] = u(e);        cout<<numbers[i]<<" ";    }    cout<<endl;    int k = 5;    int *result = FindMinK(numbers, n, k);    cout<<"minimal "<<k<<" numbers: ";    for (int i = 0; i < k; ++i) {        cout << result[i] << " ";    }    cout<<endl;    delete []result;    return 0;}

运行结果

original numbers: -3 74 61 72 38 88 12 31 36 64 minimal 5 numbers: 38 36 31 -3 12 Process finished with exit code 0

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