bfprt算法及其相关 找到无序数组中最小的K个数 【题目】 给定一个无序的整型数组arr，找到其中最小的k个数。 【要求】 如果数组arr的长度为N，排序之后自然可以得到最小的k个数，此时时间复杂度为排序的时间复杂度即O(N*logN)。本题要求读者实现时间复杂度O(N*logK)和O(N)的方法。 利用堆：     public 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 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 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;         }     }     public void swap(int[] arr, int index1, int index2) {         int tmp = arr[index1];         arr[index1] = arr[index2];         arr[index2] = tmp;     } 利用bfprt算法：     public 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 int getMinKthByBFPRT(int[] arr, int K) {         int[] copyArr = copyArray(arr);         return select(copyArr, 0, copyArr.length - 1, K - 1);     }     public 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 int select(int[] arr, int begin, int end, int i) {         if (begin == end) {             return arr[begin];         }         int pivot = medianOfMedians(arr, begin, end);         int[] pivotRange = partition(arr, begin, end, pivot);         if (i >= pivotRange[0] && i <= pivotRange[1]) {             return arr[i];         } else if (i < pivotRange[0]) {             return select(arr, begin, pivotRange[0] - 1, i);         } else {             return select(arr, pivotRange[1] + 1, end, i);         }     }     public int medianOfMedians(int[] arr, int begin, int end) {         int num = end - begin + 1;         int offset = num % 5 == 0 ? 0 : 1;         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));         }         return select(mArr, 0, mArr.length - 1, mArr.length / 2);     }     public 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 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];     }     public 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 void swap(int[] arr, int index1, int index2) {         int tmp = arr[index1];         arr[index1] = arr[index2];         arr[index2] = tmp;     }