Find the smallest number of k in an array

Title: 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,.

 Packagetest;Importjava.util.ArrayList;ImportJava.util.Comparator;ImportJava.util.PriorityQueue;Importorg.junit.Test; Public classgetleastnumbers_solution {/*** Based on priority queue, time complexity is O (NLOGK) * *@paramInput *@paramK *@return     */     PublicArraylist<integer>Getleastnumbers_solutionpriorityquene (int[] Input,intk) {ArrayList<Integer> result =NewArraylist<integer>(); if(Input = =NULL|| Input.length = = 0 | | K <= 0 | | K >input.length)returnresult; Priorityqueue<Integer> maxheap =NewPriorityqueue<integer>(                NewComparator<integer>() {@Override Public intCompare (integer O1, integer o2) {returnO2.compareto (O1);        }                });  for(inti = 0; I < K; i++) {Maxheap.offer (input[i]); }         for(intj = k; J < Input.length; J + +) {            if(Input[j] <Maxheap.peek ())                {Maxheap.poll ();            Maxheap.offer (Input[j]); }        }         for(integer integer:maxheap) {result.add (integer); }        returnresult; }    /*** Based on heap ordering, time complexity is O (NLOGK) * *@paramInput *@paramK *@return     */     PublicArraylist<integer> Getleastnumbers_solutionmaxheap (int[] Input,intk) {ArrayList<Integer> result =NewArraylist<integer>(); if(Input = =NULL|| Input.length = = 0 | | K <= 0 | | K >input.length)returnresult; //Build Maximum HeapBuiltmaxheap (input,k-1);  for(inti = k; i < input.length; i++) {            //the number after the K bit of the array is smaller than the heap top            if(Input[k] < input[0]) {input[0] =Input[k]; //Adjustment HeapBuiltmaxheap (Input, k-1); }        }         for(inti = 0; I < K; i++) {Result.add (input[i]); }        returnresult; }    /*** Build and adjust the maximum heap *@paramA *@paramLastIndex*/     Public voidBuiltmaxheap (int[]a,intLastIndex) {        intParentindex = ((lastIndex-1) >> 1); //start from the parent node of the last node         for(inti=parentindex;i>=0;i--){            //child nodes exist             while(i*2+1<=LastIndex) {                intLeftindex = i*2+1; intRightindex = i*2+2; intBiggerindex =Leftindex; //There is a right node.                if(Rightindex <=LastIndex) {                    if(A[rightindex] >A[biggerindex]) {Biggerindex=Rightindex; }                }                //the maximum node in a child node is greater than the parent node                if(A[biggerindex] >A[i])                    {swap (A,i,biggerindex); I=Biggerindex; }Else{                     Break; }            }        }    }                    /*** Based on the partition function, the time complexity is O (n), the original array has been modified *@paramInput *@paramK *@return     */     PublicArraylist<integer> Getleastnumbers_solutionpartition (int[] Input,intk) {ArrayList<Integer> result =NewArraylist<integer>(); if(Input = =NULL|| Input.length = = 0 | | K <= 0 | | K >input.length)returnresult; intleft = 0; intright = Input.length-1; intindex = 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(inti = 0; I < K; i++) {Result.add (input[i]); }        returnresult; }            /*** Partition Function *@paramA *@paramLeft *@paramRight *@return     */     Public intPartitionint[] A,intLeftintRight ) {         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); }        }        returnLeft ; }     Public voidSwapint[] A,intIintj) {intTMP =A[i]; A[i]=A[j]; A[J]=tmp; } @Test Public voidtestgetleastnumbers_solution () {int[] A = {4, 5, 1, 6, 2, 7, 3, 8 }; intK = 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 ()); }}

In addition to priority queueing, time complexity O (nlogk), heap ordering, time complexity O (NLOGK), partition function, time complexity is O (n) solution, and the solution based on bubble sorting time complexity is (NK).

Find the smallest number of k in an array