求数组中最小的k个数

Posted 2020-10-13 Tongkey

题目：输入n个整数，找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字，则最小的4个数字是1,2,3,4,。

package test;

import java.util.ArrayList;
import java.util.Comparator;
import java.util.PriorityQueue;

import org.junit.Test;

public class GetLeastNumbers_Solution {
    /**
     * 基于优先队列，时间复杂度为o(nlogk)
     * 
     * @param input
     * @param k
     * @return
     */
    public ArrayList<Integer> GetLeastNumbers_SolutionPriorityQuene(
            int[] input, int k) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (input == null || input.length == 0 || k <= 0 || k > input.length)
            return result;
        PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(
                new Comparator<Integer>() {
                    @Override
                    public int compare(Integer o1, Integer o2) {
                        return o2.compareTo(o1);
                    }
                });
        for (int i = 0; i < k; i++) {
            maxHeap.offer(input[i]);
        }
        for (int j = k; j < input.length; j++) {
            if (input[j] < maxHeap.peek()) {
                maxHeap.poll();
                maxHeap.offer(input[j]);
            }
        }
        for (Integer integer : maxHeap) {
            result.add(integer);
        }

        return result;
    }

    /**
     * 基于堆排序,时间复杂度为o(nlogk)
     * 
     * @param input
     * @param k
     * @return
     */
    public ArrayList<Integer> GetLeastNumbers_SolutionMaxHeap(int[] input, int k) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (input == null || input.length == 0 || k <= 0 || k > input.length)
            return result;
        //构建最大堆
        builtMaxHeap(input,k-1);
        for (int i = k; i < input.length; i++) {
            //数组k位后的数字比堆顶小
            if (input[k] < input[0]) {
                input[0] = input[k];
                //调整堆
                builtMaxHeap(input, k - 1);
            }
        }
        for (int i = 0; i < k; i++) {
            result.add(input[i]);
        }
        return result;
    }

    /**
     * 构建、调整最大堆
     * @param a
     * @param lastIndex
     */
    public void builtMaxHeap(int[]a,int lastIndex){
        int parentIndex = ((lastIndex-1) >> 1);
        //从最后一个节点的父节点开始
        for(int i=parentIndex;i>=0;i--){
            //存在子节点
            while (i*2+1<=lastIndex){
                int leftIndex = i*2+1;
                int rightIndex = i*2+2;
                int biggerIndex = leftIndex;
                //存在右结点
                if (rightIndex <= lastIndex){
                    if(a[rightIndex] > a[biggerIndex]){
                        biggerIndex = rightIndex;
                    }
                }
                //子节点中最大节点大于父节点
                if (a[biggerIndex] > a[i]){
                    swap(a,i,biggerIndex);
                    i = biggerIndex;
                }else{
                    break;
                }
            }
        }
    }
        
    
    
    /**
     * 基于Partition函数,时间复杂度为o(n),原数组已被修改
     * 
     * @param input
     * @param k
     * @return
     */
    public ArrayList<Integer> GetLeastNumbers_SolutionPartition(int[] input,
            int k) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (input == null || input.length == 0 || k <= 0 || k > input.length)
            return result;

        int left = 0;
        int right = input.length - 1;
        int index = partition(input, 0, right);

        while (index != k - 1) {
            if (index > k - 1) {
                right = index - 1;
                index = partition(input, left, right);
            } else {
                left = index + 1;
                index = partition(input, left, right);
            }
        }
        for (int i = 0; i < k; i++) {
            result.add(input[i]);
        }

        return result;
    }
    
    
    /**
     * partition函数
     * @param a
     * @param left
     * @param right
     * @return
     */
    public int partition(int[] a, int left, int right) {
        while (left < right) {
            while (left < right && a[left] <= a[right]) {
                right--;
            }
            if (left < right) {
                swap(a, left, right);
            }
            while (left < right && a[left] <= a[right]) {
                left++;
            }
            if (left < right) {
                swap(a, left, right);
            }

        }
        return left;
    }

    public void swap(int[] a, int i, int j) {
        int tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }

    @Test
    public void testGetLeastNumbers_Solution() {
        int[] a = { 4, 5, 1, 6, 2, 7, 3, 8 };
        int k = 4;
        ArrayList<Integer> list = GetLeastNumbers_SolutionPartition(a, k);
        System.out.println(list.toString());

        ArrayList<Integer> list2 = GetLeastNumbers_SolutionPriorityQuene(a, k);
        System.out.println(list2.toString());
        
        ArrayList<Integer> list3 = GetLeastNumbers_SolutionMaxHeap(a, k);
        System.out.println(list3.toString());
        
    }
}

 除了基于优先队列，时间复杂度为O（nlogk）、堆排序，时间复杂度为O（nlogk）、partition函数，时间复杂度为O（n）的解法之外，还有基于冒泡排序的解法时间复杂度为（nk）。