算法习题之bfprt算法及蓄水池算法

bfprt算法与改写快排的方法的唯一的不同就是，bfprt算法对于找的随机数讲究
1.3个数一组7个数一组也可以收敛到O(N)，因为是5个人发明的，所以是5个一组

习题1 在无序数组中求第K小的数 1）改写快排的方法 2）bfprt算法

public static class MaxHeapComparator implements Comparator {

		@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 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");
	}

习题2 给定一个无序数组arr中，长度为N，给定一个正数k，返回top k个最大的数 不同时间复杂度三个方法：1）O(NlogN) 2）O(N + KlogN) 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"));
	}

习题3 蓄水池算法 假设有一个源源吐出不同球的机器，只有装下10个球的袋子，每一个吐出的球，要么放入袋子，要么永远扔掉 如何做到机器吐出每一个球之后，所有吐出的球都等概率被放进袋子里

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]);
		}

	}