TopK问题——求数组中第K小的数

public class Problem_01_FindMinKNums {// O(N*logK)public static int[] getMinKNumsByHeap(int[] arr, int k) {if (k < 1 || k > arr.length) {return arr;}int[] kHeap = new int[k];for (int i = 0; i != k; i++) {heapInsert(kHeap, arr[i], i);}for (int i = k; i != arr.length; i++) {if (arr[i] < kHeap[0]) {kHeap[0] = arr[i];heapify(kHeap, 0, k);}}return kHeap;}public static void heapInsert(int[] arr, int value, int index) {arr[index] = value;while (index != 0) {int parent = (index - 1) / 2;if (arr[parent] < arr[index]) {swap(arr, parent, index);index = parent;} else {break;}}}public static void heapify(int[] arr, int index, int heapSize) {int left = index * 2 + 1;int right = index * 2 + 2;int largest = index;while (left < heapSize) {if (arr[left] > arr[index]) {largest = left;}if (right < heapSize && arr[right] > arr[largest]) {largest = right;}if (largest != index) {swap(arr, largest, index);} else {break;}index = largest;left = index * 2 + 1;right = index * 2 + 2;}}// O(N)public static int[] getMinKNumsByBFPRT(int[] arr, int k) {if (k < 1 || k > arr.length) {return arr;}int minKth = getMinKthByBFPRT(arr, k);int[] res = new int[k];int index = 0;for (int i = 0; i != arr.length; i++) {if (arr[i] < minKth) {res[index++] = arr[i];}}for (; index != res.length; index++) {res[index] = minKth;}return res;}public static int getMinKthByBFPRT(int[] arr, int K) {int[] copyArr = copyArray(arr);return select(copyArr, 0, copyArr.length - 1, K - 1);}public static int[] copyArray(int[] arr) {int[] res = new int[arr.length];for (int i = 0; i != res.length; i++) {res[i] = arr[i];}return res;}public static int select(int[] arr, int begin, int end, int i) {if (begin == end) {return arr[begin];}int pivot = medianOfMedians(arr, begin, end);  //找中位数组的中位数int[] pivotRange = partition(arr, begin, end, pivot);  //为什么返回数组，因为要找等于的范围，因此有两个下标if (i >= pivotRange[0] && i <= pivotRange[1]) {  //如果命中，则直接返回return arr[i];} else if (i < pivotRange[0]) {   //没命中，左边递归或右边递归return select(arr, begin, pivotRange[0] - 1, i);} else {return select(arr, pivotRange[1] + 1, end, i);}}public static int medianOfMedians(int[] arr, int begin, int end) {int num = end - begin + 1;int offset = num % 5 == 0 ? 0 : 1;  //不满5个数单独成一组int[] mArr = new int[num / 5 + offset];for (int i = 0; i < mArr.length; i++) {int beginI = begin + i * 5;int endI = beginI + 4;mArr[i] = getMedian(arr, beginI, Math.min(end, endI)); //保存每个组的中位数}return select(mArr, 0, mArr.length - 1, mArr.length / 2);  //返回中位数组中的中位数}public static int[] partition(int[] arr, int begin, int end, int pivotValue) {int small = begin - 1;int cur = begin;int big = end + 1;while (cur != big) {if (arr[cur] < pivotValue) {swap(arr, ++small, cur++);} else if (arr[cur] > pivotValue) {swap(arr, cur, --big);} else {cur++;}}int[] range = new int[2];range[0] = small + 1;range[1] = big - 1;return range;}public static int getMedian(int[] arr, int begin, int end) {insertionSort(arr, begin, end);int sum = end + begin;int mid = (sum / 2) + (sum % 2);return arr[mid];}//5个数进行排序，选了一个常数项低的插入排序public static void insertionSort(int[] arr, int begin, int end) {for (int i = begin + 1; i != end + 1; i++) {for (int j = i; j != begin; j--) {if (arr[j - 1] > arr[j]) {swap(arr, j - 1, j);} else {break;}}}}public static void swap(int[] arr, int index1, int index2) {int tmp = arr[index1];arr[index1] = arr[index2];arr[index2] = tmp;}public static void printArray(int[] arr) {for (int i = 0; i != arr.length; i++) {System.out.print(arr[i] + " ");}System.out.println();}public static void main(String[] args) {int[] arr = { 6, 9, 1, 3, 1, 2, 2, 5, 6, 1, 3, 5, 9, 7, 2, 5, 6, 1, 9 };// sorted : { 1, 1, 1, 1, 2, 2, 2, 3, 3, 5, 5, 5, 6, 6, 6, 7, 9, 9, 9 }printArray(getMinKNumsByHeap(arr, 10));printArray(getMinKNumsByBFPRT(arr, 10));}}