BFPRT算法之解决Top-K问题

旧城等待， 2022-05-15 09:57 184阅读 0赞

# 一、[概要][Link 1] #

**在一大堆数中求其前k大或前k小的问题，简称TOP-K问题**。而目前解决TOP-K问题最有效的算法即是BFPRT算法，其又称为中位数的中位数算法，该算法由Blum、Floyd、Pratt、Rivest、Tarjan提出，最坏时间复杂度为O(n)。

# 二、算法实现 #

BFPRT算法步骤如下：  
（1）：选取主元；  
  （1.1）：将n个元素划分为⌊n5⌋个组，每组5个元素，若有剩余，舍去；  
  （1.2）：使用插入排序找到⌊n5⌋个组中每一组的中位数；  
  （1.3）：对于（1.2）中找到的所有中位数，调用BFPRT算法求出它们的中位数，作为主元；  
（2）：以（1.3）选取的主元为分界点，把小于主元的放在左边，大于主元的放在右边；  
（3）：判断主元的位置与k的大小，有选择的对左边或右边递归。

/** * 在一大堆数中求其前k大或前k小的问题 * @author superman * */
    public class BFPRT { 
    
        // 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);
            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 static 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 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];
        }
    
        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));
    
        }
    
    }

[Link 1]: https://blog.csdn.net/laojiu_/article/details/54986553