最小的K个数
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。
思路
最简单的方法就是先排序,然后在遍历输出最小的K个数,方法简单粗暴。
代码
import java.util.ArrayList;
public class Solution {public ArrayList GetLeastNumbers_Solution(int [] input, int k) {ArrayList returnList &#61; new ArrayList<>();if(input &#61;&#61; null || input.length &#61;&#61;0 || k <1 || k>input.length){return returnList;}int[] ks &#61; new int[k];for(int i &#61;0;i&#61;0;i--){toHeap(ks,i,ks.length);}for(int i &#61;k;i input[index]){index &#61; left;}while(right input[index]){index &#61; right;}if(index !&#61;start){int temp &#61; input[start];input[start] &#61; input[index];input[index] &#61; temp;toHeap(input,index,end);}}
}
代码1 找一个放置的空间
public ArrayList GetLeastNumbers_Solution(int [] input, int k) {List list &#61; new ArrayList();if (k>input.length || k&#61;&#61;0) {return (ArrayList)list;}for (int i &#61; 0; i }
代码2 PriorityQueue
public static void main(String[] args) {int [] input &#61; {4,5,1,6,2,7,3,8};System.out.println(GetLeastNumbers_Solution(input,4));
}public static ArrayList GetLeastNumbers_Solution(int [] input, int k) {ArrayList list &#61; new ArrayList<>();if (input &#61;&#61; null || k <&#61; 0 || k > input.length || input.length&#61;&#61;0) {return list;}PriorityQueue pq &#61; new PriorityQueue<>(k, new Comparator(){&#64;Overridepublic int compare(Integer o1, Integer o2){return o2 - o1;}});for(int i&#61;0; i}
代码3
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Arrays;
public class Solution {
public ArrayList GetLeastNumbers_Solution(int[] input, int k) {ArrayList list &#61; new ArrayList<>();if (input &#61;&#61; null || k <&#61; 0 || k > input.length) {return list;}int[] kArray &#61; Arrays.copyOfRange(input, 0, k);//1.构建大顶堆for(int i&#61;kArray.length / 2 - 1;i>&#61;0;i--){//从第一个非叶子节点开始&#xff0c;下次第二个&#xff0c;依次调整构建大根堆adjustHeap(kArray,i,kArray.length);}//2.依次来判断&#xff0c;有小于堆顶的那就替换掉继续调整一个堆for (int i &#61; k; i &#61; 0; i--) {list.add(kArray[i]);}return list;}public static void adjustHeap(int []arr,int i,int length){ //从i节点开始调整&#xff0c;int temp &#61; arr[i];//先取出当前元素ifor(int k &#61; i*2&#43;1;ktemp){//如果子节点大于父节点&#xff0c;将子节点值赋给父节点&#xff08;不用进行交换&#xff09;arr[i] &#61; arr[k];i &#61; k;}else{break;}}arr[i] &#61; temp;//将temp值放到最终的位置}
代码4
用前K个数建立最大堆,每次用堆顶元素和n-k中各个元素比较,如果堆顶元素较大,则互换位置,然后调整堆,使之重新成为最大堆。时间复杂度
O&#xff08;n*logk&#xff09;
import java.util.ArrayList;/*** &#64;Auther: liuhaidong* Data: 2020/1/14 0014、14:05* Description:* &#64;version: 1.0*/
public class Test17 {public static void main(String[] args) {int [] arr &#61;{4,5,1,6,2,7,3,8};System.out.println(GetLeastNumbers_Solution(arr,4));}public static ArrayList GetLeastNumbers_Solution(int [] ins, int k) {if(ins &#61;&#61; null || k>ins.length){return new ArrayList<>();}int[] ks &#61; new int[k];// 最先遍历的k个数放入数组中 o(k)for (int i &#61; 0; i &#61; 0; i--) {maxHeap(ks, ks.length, i);}//n-k个数和前面和k中最大数比较for (int i &#61;k; i ins[i]){ks[0]&#61;ins[i];//调整堆,堆顶被替换了,加入被替换的值非常小,会一直下沉到叶子节点.maxHeap(ks,ks.length,0);}}ArrayList lists &#61; new ArrayList<>();// 输出最小的K个数for (int i &#61; 0; i array[largest]) {largest &#61; left;}//判断有没有右节点,如若有则largest和右节点比较,注意largest有可能是left也有可能是indexif (right array[largest]) {largest &#61; right;}if (index !&#61; largest) {int temp &#61; array[index];array[index] &#61; array[largest];array[largest] &#61; temp;//被替换的largest节点所在的堆,需要重新调整,使小值/大值一直下沉maxHeap(array, heapSize, largest);}}
}
关键代码
int[] ks &#61; new int[k];// 最先遍历的k个数放入数组中 o(k)for (int i &#61; 0; i &#61; 0; i--) {maxHeap(ks, ks.length, i);}
这个代码相当于构建堆
参考网站 bilibili.com/video/av47196993/