18. Algorithm exercises: bfprt algorithm and reservoir algorithm

The only difference between the bfprt algorithm and the method of rewriting quick sort is that the bfprt algorithm requires a group of 1.3 numbers and a group of 7 numbers to find random numbers. It can also converge to O(N). Because it was invented by 5 people, it is
group of 5

Exercise 1: Find the Kth smallest number in an unordered array 1) Rewrite the quick sort method 2) bfprt algorithm

public static class MaxHeapComparator implements Comparator<Integer> {

		@Override
		public int compare(Integer o1, Integer o2) {
			return o2 - o1;
		}

	}

	// 利用大根堆，时间复杂度O(N*logK)
	public static int minKth1(int[] arr, int k) {
		PriorityQueue<Integer> maxHeap = new PriorityQueue<>(new MaxHeapComparator());
		for (int i = 0; i < k; i++) {
			maxHeap.add(arr[i]);
		}
		for (int i = k; i < arr.length; i++) {
			if (arr[i] < maxHeap.peek()) {
				maxHeap.poll();
				maxHeap.add(arr[i]);
			}
		}
		return maxHeap.peek();
	}

	// 改写快排，时间复杂度O(N)
	// k >= 1
	public static int minKth2(int[] array, int k) {
		int[] arr = copyArray(array);
		return process2(arr, 0, arr.length - 1, k - 1);
	}

	public static int[] copyArray(int[] arr) {
		int[] ans = new int[arr.length];
		for (int i = 0; i != ans.length; i++) {
			ans[i] = arr[i];
		}
		return ans;
	}

	// arr 第k小的数
	// process2(arr, 0, N-1, k-1)
	// arr[L..R]  范围上，如果排序的话(不是真的去排序)，找位于index的数
	// index [L..R]
	public static int process2(int[] arr, int L, int R, int index) {
		if (L == R) { // L = =R ==INDEX
			return arr[L];
		}
		// 不止一个数  L +  [0, R -L]
		int pivot = arr[L + (int) (Math.random() * (R - L + 1))];
		int[] range = partition(arr, L, R, pivot);
		if (index >= range[0] && index <= range[1]) {
			return arr[index];
		} else if (index < range[0]) {
			return process2(arr, L, range[0] - 1, index);
		} else {
			return process2(arr, range[1] + 1, R, index);
		}
	}

	public static int[] partition(int[] arr, int L, int R, int pivot) {
		int less = L - 1;
		int more = R + 1;
		int cur = L;
		while (cur < more) {
			if (arr[cur] < pivot) {
				swap(arr, ++less, cur++);
			} else if (arr[cur] > pivot) {
				swap(arr, cur, --more);
			} else {
				cur++;
			}
		}
		return new int[] { less + 1, more - 1 };
	}

	public static void swap(int[] arr, int i1, int i2) {
		int tmp = arr[i1];
		arr[i1] = arr[i2];
		arr[i2] = tmp;
	}

	// 利用bfprt算法，时间复杂度O(N)
	public static int minKth3(int[] array, int k) {
		int[] arr = copyArray(array);
		return bfprt(arr, 0, arr.length - 1, k - 1);
	}

	// arr[L..R]  如果排序的话，位于index位置的数，是什么，返回
	public static int bfprt(int[] arr, int L, int R, int index) {
		if (L == R) {
			return arr[L];
		}
		// L...R  每五个数一组
		// 每一个小组内部排好序
		// 小组的中位数组成新数组
		// 这个新数组的中位数返回
		int pivot = medianOfMedians(arr, L, R);
		int[] range = partition(arr, L, R, pivot);
		if (index >= range[0] && index <= range[1]) {
			return arr[index];
		} else if (index < range[0]) {
			return bfprt(arr, L, range[0] - 1, index);
		} else {
			return bfprt(arr, range[1] + 1, R, index);
		}
	}

	// arr[L...R]  五个数一组
	// 每个小组内部排序
	// 每个小组中位数领出来，组成marr
	// marr中的中位数，返回
	public static int medianOfMedians(int[] arr, int L, int R) {
		int size = R - L + 1;
		int offset = size % 5 == 0 ? 0 : 1;
		int[] mArr = new int[size / 5 + offset];
		for (int team = 0; team < mArr.length; team++) {
			int teamFirst = L + team * 5;
			// L ... L + 4
			// L +5 ... L +9
			// L +10....L+14
			mArr[team] = getMedian(arr, teamFirst, Math.min(R, teamFirst + 4));
		}
		// marr中，找到中位数
		// marr(0, marr.len - 1,  mArr.length / 2 )
		return bfprt(mArr, 0, mArr.length - 1, mArr.length / 2);
	}

	public static int getMedian(int[] arr, int L, int R) {
		insertionSort(arr, L, R);
		return arr[(L + R) / 2];
	}

	public static void insertionSort(int[] arr, int L, int R) {
		for (int i = L + 1; i <= R; i++) {
			for (int j = i - 1; j >= L && arr[j] > arr[j + 1]; j--) {
				swap(arr, j, j + 1);
			}
		}
	}

	// for test
	public static int[] generateRandomArray(int maxSize, int maxValue) {
		int[] arr = new int[(int) (Math.random() * maxSize) + 1];
		for (int i = 0; i < arr.length; i++) {
			arr[i] = (int) (Math.random() * (maxValue + 1));
		}
		return arr;
	}

	public static void main(String[] args) {
		int testTime = 1000000;
		int maxSize = 100;
		int maxValue = 100;
		System.out.println("test begin");
		for (int i = 0; i < testTime; i++) {
			int[] arr = generateRandomArray(maxSize, maxValue);
			int k = (int) (Math.random() * arr.length) + 1;
			int ans1 = minKth1(arr, k);
			int ans2 = minKth2(arr, k);
			int ans3 = minKth3(arr, k);
			if (ans1 != ans2 || ans2 != ans3) {
				System.out.println("Oops!");
			}
		}
		System.out.println("test finish");
	}

Exercise 2: Given an unordered array arr with length N and a positive number k, return the top k largest numbers. Three methods with different time complexity: 1) O(N logN) 2) O(N + K logN) 3) O(n + k*logk)

// 时间复杂度O(N*logN)
	// 排序+收集
	public static int[] maxTopK1(int[] arr, int k) {
		if (arr == null || arr.length == 0) {
			return new int[0];
		}
		int N = arr.length;
		k = Math.min(N, k);
		Arrays.sort(arr);
		int[] ans = new int[k];
		for (int i = N - 1, j = 0; j < k; i--, j++) {
			ans[j] = arr[i];
		}
		return ans;
	}

	// 方法二，时间复杂度O(N + K*logN)
	// 解释：堆
	public static int[] maxTopK2(int[] arr, int k) {
		if (arr == null || arr.length == 0) {
			return new int[0];
		}
		int N = arr.length;
		k = Math.min(N, k);
		// 从底向上建堆，时间复杂度O(N)
		for (int i = N - 1; i >= 0; i--) {
			heapify(arr, i, N);
		}
		// 只把前K个数放在arr末尾，然后收集，O(K*logN)
		int heapSize = N;
		swap(arr, 0, --heapSize);
		int count = 1;
		while (heapSize > 0 && count < k) {
			heapify(arr, 0, heapSize);
			swap(arr, 0, --heapSize);
			count++;
		}
		int[] ans = new int[k];
		for (int i = N - 1, j = 0; j < k; i--, j++) {
			ans[j] = arr[i];
		}
		return ans;
	}

	public static void heapInsert(int[] arr, int index) {
		while (arr[index] > arr[(index - 1) / 2]) {
			swap(arr, index, (index - 1) / 2);
			index = (index - 1) / 2;
		}
	}

	public static void heapify(int[] arr, int index, int heapSize) {
		int left = index * 2 + 1;
		while (left < heapSize) {
			int largest = left + 1 < heapSize && arr[left + 1] > arr[left] ? left + 1 : left;
			largest = arr[largest] > arr[index] ? largest : index;
			if (largest == index) {
				break;
			}
			swap(arr, largest, index);
			index = largest;
			left = index * 2 + 1;
		}
	}

	public static void swap(int[] arr, int i, int j) {
		int tmp = arr[i];
		arr[i] = arr[j];
		arr[j] = tmp;
	}

	// 方法三，时间复杂度O(n + k * logk)
	public static int[] maxTopK3(int[] arr, int k) {
		if (arr == null || arr.length == 0) {
			return new int[0];
		}
		int N = arr.length;
		k = Math.min(N, k);
		// O(N)
		int num = minKth(arr, N - k);
		int[] ans = new int[k];
		int index = 0;
		for (int i = 0; i < N; i++) {
			if (arr[i] > num) {
				ans[index++] = arr[i];
			}
		}
		for (; index < k; index++) {
			ans[index] = num;
		}
		// O(k*logk)
		Arrays.sort(ans);
		for (int L = 0, R = k - 1; L < R; L++, R--) {
			swap(ans, L, R);
		}
		return ans;
	}

	// 时间复杂度O(N)
	public static int minKth(int[] arr, int index) {
		int L = 0;
		int R = arr.length - 1;
		int pivot = 0;
		int[] range = null;
		while (L < R) {
			pivot = arr[L + (int) (Math.random() * (R - L + 1))];
			range = partition(arr, L, R, pivot);
			if (index < range[0]) {
				R = range[0] - 1;
			} else if (index > range[1]) {
				L = range[1] + 1;
			} else {
				return pivot;
			}
		}
		return arr[L];
	}

	public static int[] partition(int[] arr, int L, int R, int pivot) {
		int less = L - 1;
		int more = R + 1;
		int cur = L;
		while (cur < more) {
			if (arr[cur] < pivot) {
				swap(arr, ++less, cur++);
			} else if (arr[cur] > pivot) {
				swap(arr, cur, --more);
			} else {
				cur++;
			}
		}
		return new int[] { less + 1, more - 1 };
	}

	// for test
	public static int[] generateRandomArray(int maxSize, int maxValue) {
		int[] arr = new int[(int) ((maxSize + 1) * Math.random())];
		for (int i = 0; i < arr.length; i++) {
			// [-? , +?]
			arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random());
		}
		return arr;
	}

	// for test
	public static int[] copyArray(int[] arr) {
		if (arr == null) {
			return null;
		}
		int[] res = new int[arr.length];
		for (int i = 0; i < arr.length; i++) {
			res[i] = arr[i];
		}
		return res;
	}

	// for test
	public static boolean isEqual(int[] arr1, int[] arr2) {
		if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) {
			return false;
		}
		if (arr1 == null && arr2 == null) {
			return true;
		}
		if (arr1.length != arr2.length) {
			return false;
		}
		for (int i = 0; i < arr1.length; i++) {
			if (arr1[i] != arr2[i]) {
				return false;
			}
		}
		return true;
	}

	// for test
	public static void printArray(int[] arr) {
		if (arr == null) {
			return;
		}
		for (int i = 0; i < arr.length; i++) {
			System.out.print(arr[i] + " ");
		}
		System.out.println();
	}

	// 生成随机数组测试
	public static void main(String[] args) {
		int testTime = 500000;
		int maxSize = 100;
		int maxValue = 100;
		boolean pass = true;
		System.out.println("测试开始，没有打印出错信息说明测试通过");
		for (int i = 0; i < testTime; i++) {
			int k = (int) (Math.random() * maxSize) + 1;
			int[] arr = generateRandomArray(maxSize, maxValue);

			int[] arr1 = copyArray(arr);
			int[] arr2 = copyArray(arr);
			int[] arr3 = copyArray(arr);

			int[] ans1 = maxTopK1(arr1, k);
			int[] ans2 = maxTopK2(arr2, k);
			int[] ans3 = maxTopK3(arr3, k);
			if (!isEqual(ans1, ans2) || !isEqual(ans1, ans3)) {
				pass = false;
				System.out.println("出错了！");
				printArray(ans1);
				printArray(ans2);
				printArray(ans3);
				break;
			}
		}
		System.out.println("测试结束了，测试了" + testTime + "组，是否所有测试用例都通过？" + (pass ? "是" : "否"));
	}

Exercise 3 Reservoir Algorithm Suppose there is a machine that spits out different balls one after another. There is only a bag that can hold 10 balls. Every ball spit out must be put into the bag or thrown away forever. How to make sure that after the machine spits out each ball, All spit out balls have equal probability of being put into the bag

public static class RandomBox {
		private int[] bag;
		private int N;
		private int count;

		public RandomBox(int capacity) {
			bag = new int[capacity];
			N = capacity;
			count = 0;
		}

		private int rand(int max) {
			return (int) (Math.random() * max) + 1;
		}

		public void add(int num) {
			count++;
			if (count <= N) {
				bag[count - 1] = num;
			} else {
				if (rand(count) <= N) {
					bag[rand(N) - 1] = num;
				}
			}
		}

		public int[] choices() {
			int[] ans = new int[N];
			for (int i = 0; i < N; i++) {
				ans[i] = bag[i];
			}
			return ans;
		}

	}

	// 请等概率返回1~i中的一个数字
	public static int random(int i) {
		return (int) (Math.random() * i) + 1;
	}

	public static void main(String[] args) {
		System.out.println("hello");
		int test = 10000;
		int ballNum = 17;
		int[] count = new int[ballNum + 1];
		for (int i = 0; i < test; i++) {
			int[] bag = new int[10];
			int bagi = 0;
			for (int num = 1; num <= ballNum; num++) {
				if (num <= 10) {
					bag[bagi++] = num;
				} else { // num > 10
					if (random(num) <= 10) { // 一定要把num球入袋子
						bagi = (int) (Math.random() * 10);
						bag[bagi] = num;
					}
				}

			}
			for (int num : bag) {
				count[num]++;
			}
		}
		for (int i = 0; i <= ballNum; i++) {
			System.out.println(count[i]);
		}

		System.out.println("hello");
		int all = 100;
		int choose = 10;
		int testTimes = 50000;
		int[] counts = new int[all + 1];
		for (int i = 0; i < testTimes; i++) {
			RandomBox box = new RandomBox(choose);
			for (int num = 1; num <= all; num++) {
				box.add(num);
			}
			int[] ans = box.choices();
			for (int j = 0; j < ans.length; j++) {
				counts[ans[j]]++;
			}
		}

		for (int i = 0; i < counts.length; i++) {
			System.out.println(i + " times : " + counts[i]);
		}

	}