百度面试题 20个数组 每个500个元素 升序排列 找出前500的数

一道面试题

有20个数组，每个数组有500个元素，升序排列，现在在这20*500个数中找出排名前500的数。

转载请注明出处 http://www.alijava.com/arrays-sort/

方法思路分析

笔者想到3种方法，有更好的方法欢迎评论留言。

1. 直接暴力求解，将20个数组合并，然后排序，取出500个数

2. 使用归并。 对于排好序的 序列，我们要注意使用归并。
   先将第1个和第2个归并，得到500个数据（没必要归并得到总的1000个数）。然后再加结果和第3个归并，得到500个数据，再与第4个归并，等等。

3. 利用堆。保持一个20的堆，然后先将每个数组的第1个数入堆。

   20个元素的堆一直保持容量为20个，20个数组的最小元素可以将20个数组的第0个元素入堆，最小堆的性质，顶点为最小值。这时候得到了500个结果里的第0个结果。然后再把下一个元素入20个元素的堆，堆插入的时候会保持性质不变，最小元素依然在顶点。再取出20个元素的顶点，得到500个结果里的第1个结果。

   假设 [1,3,4,5,6] [2,3,4,5,6] [3,4,5,6,7]
   最小的是比较 1 2 3 得到1
   次小的是将刚才的1替换为它后面的元素3, 再加上刚才的元素2 3，得到2

   注意这儿需要保持数来自于哪个数组，以及其在数组里的位置

实力代码

方法1

// 方法1，直接将20个数组合并，排序，然后取前500个
// 复杂度是 10000log(10000)
Integer[] allData = new Integer[20 * 500];
int a = 0;
for (int ii = 0; ii < rowSize; ii++) {
for (int i = 0; i < columnSize; i++) {
allData[a++] = data[ii][i];
}
}
Arrays.sort(allData);
Integer[] result1 = Arrays.copyOfRange(allData, 0, columnSize);
System.out.println("method 1 result:        " + Arrays.toString(result1));

方法2

先定义一个 merge 函数，从两个数组里面得到前500个数据。

/**
 * 
 * @param first
 *            数组1
 * @param second
 *            数组2
 * @param n
 *            数组的长度，假定 first 和 second 的长度均为 n
 * @return 返回一个长度为 n 的数组
 */
private static Integer[] merge(Integer[] first, Integer[] second, int n) {

	// 构建一个数组，这是归并排序的缺点，需要额外空间
	Integer[] temp = new Integer[n];

	int f = 0;
	int s = 0;

	int i = 0;

	while (i != n && f <= n && s <= n) {
		if (first[f] < second[s]) {
			temp[i++] = first[f++];
		} else {
			temp[i++] = second[s++];
		}
	}

	return temp;

}

下来是总的归并代码

// 方法2，归并，先归并前两个数组，取前500个数据。将得到的结果再与第3个归并，重复
// 考虑一下,20个数组先两两归并，得到10个；10个再两两归并，得到5个。等等，直到最后1个
// 但这样归并的次数和从前往后两两归并的次数是一样的？所以还是 上述的从前往后好了
// 复杂度是 19*500
Integer[] result2 = data[0];
for (int i = 1; i < rowSize; i++) {
	result2 = merge(result2, data[i], columnSize);
}

System.out.println("method 2 result:        " + Arrays.toString(result2));

System.out.println("result1 equals result2: " + Arrays.equals(result1, result2));

方法3

定义一个最小堆，有插入和删除方法

/**
 * 
 * 最小堆和堆排序, 最小堆，顶点的元素是最小值， 根据《Java 语言程序设计 进阶篇》 p83 改写， 书上是最大堆. 堆排序
 * 将元素都存入最小堆中，从最小堆里面每次取出顶点元素
 * 
 * @author tomchen
 *
 * @param <E>
 */
class MinHeap<E extends Comparable> {

	public static <E extends Comparable> void heapSort(E[] array) {
		MinHeap<E> heap = new MinHeap<E>();
		for (int i = 0; i < array.length; i++) {
			heap.add(array[i]);
		}

		System.out.println("Debug: heap is     " + heap);

		for (int i = 0; i < array.length; i++) {
			array[i] = heap.removeTop();
		}
	}

	private ArrayList<E> data = new ArrayList<E>();

	public MinHeap() {
	}

	/**
	 * 增加一个新元素，步骤是 1. 先把元素插入到 list 的末尾 2. 比较末尾元素和它的父元素，若小于，交换两者 3.
	 * 重复上述步骤，直到到顶点位置或者子元素大于父元素 4. 不一定要遍历堆所有的元素，达到堆的性质后会提前结束
	 * 
	 * @param array
	 */
	public void add(E array) {

		data.add(array);

		int child = data.size() - 1;
		int parent = (child - 1) / 2;

		// 判断是否到达顶点
		while (child > 0) {
			// 父元素大于子元素，交换，保持父是小的
			if (data.get(parent).compareTo(array) > 0) {
				data.set(child, data.get(parent));
				data.set(parent, array);

				child = parent;
				parent = (child - 1) / 2;
			} else {
				// 已经是最小堆了，无需再比较
				break;
			}
		}
	}

	/**
	 * 删除顶点处的元素，步骤是： 1. 把末尾的元素复制到顶点处 2. 然后比较此时顶点的值和左右子树，保持最小堆的性质 3.
	 * 交换顶点和左右子树较小的值 4. 重复上述步骤，直到已经成了最小堆或者遍历完 5. 注意可能存在左子树存在，右子树不存在情况 6.
	 * 不一定要遍历堆所有的元素，达到堆的性质后会提前结束
	 * 
	 * @return 返回被删除的元素
	 */
	public E removeTop() {
		if (data.isEmpty())
			return null;

		E removed = data.get(0);

		// 因为一直交换的是最后的元素，这儿将其保存
		E last = data.get(data.size() - 1);
		data.set(0, last);
		data.remove(data.size() - 1);

		int parent = 0;
		int leftChild = parent * 2 + 1;
		int rightChild = parent * 2 + 2;

		while (leftChild <= data.size() - 1) {

			int minIndex = leftChild;
			// 右子树存在，判断左右子树哪个小，保存坐标
			// 如果不存在，那么使用左子树的坐标
			// 保存较小元素的坐标，可以省去考虑左右子树都存在，只有左存在的情况
			if (rightChild <= data.size() - 1) {
				if (data.get(rightChild).compareTo(data.get(leftChild)) < 0) {
					minIndex = rightChild;
				}
			}

			if (data.get(minIndex).compareTo(last) < 0) {
				data.set(parent, data.get(minIndex));
				data.set(minIndex, last);
				parent = minIndex;
				leftChild = parent * 2 + 1;
				rightChild = parent * 2 + 2;
			} else {
				break; // 已经达到了最小堆的性质
			}
		}

		return removed;
	}

	@Override
	public String toString() {
		return data.toString();
	}

}

为了能够记录堆里的元素来自哪个数组，以及在数组里的位置，我们使用一个内部类

class DataWithSource implements Comparable<DataWithSource> {
	// 数据
	private Integer value;
	// 来源的数组
	private Integer comeFrom;
	// 在数组中的 index
	private Integer index;

	public DataWithSource(Integer value, Integer comeFrom, Integer index) {
		this.value = value;
		this.comeFrom = comeFrom;
		this.index = index;
	}

	public Integer getComeFrom() {
		return comeFrom;
	}

	public Integer getValue() {
		return value;
	}

	public void setValue(Integer value) {
		this.value = value;
	}

	public Integer getIndex() {
		return index;
	}

	public void setIndex(Integer index) {
		this.index = index;
	}

	public int compareTo(DataWithSource o) {
		return this.value.compareTo(o.value);
	}
}

下面是方法3的代码

public static void main(String[] args) {
	Random r = new Random();

	int rowSize = 20;
	int columnSize = 500;

	// 注意 java 二维数组其实是一维数组，里面包含的也是一维数组
	Integer[][] data = new Integer[rowSize][columnSize];

	for (int ii = 0; ii < rowSize; ii++) {
		for (int i = 0; i < columnSize; i++) {
			data[ii][i] = r.nextInt(1600);
		}

		// 将500个元素排序，升序
		Arrays.sort(data[ii]);
		// System.out.println(Arrays.toString(data[ii]));
	}

	// 方法1，直接将20个数组合并，排序，然后取前500个
	// 复杂度是 10000log(10000)
	Integer[] allData = new Integer[20 * 500];
	int a = 0;
	for (int ii = 0; ii < rowSize; ii++) {
		for (int i = 0; i < columnSize; i++) {
			allData[a++] = data[ii][i];
		}
	}
	Arrays.sort(allData);
	Integer[] result1 = Arrays.copyOfRange(allData, 0, columnSize);
	System.out.println("method 1 result:        " + Arrays.toString(result1));

	// 方法2，归并，先归并前两个数组，取前500个数据。将得到的结果再与第3个归并，重复
	// 考虑一下,20个数组先两两归并，得到10个；10个再两两归并，得到5个。等等，直到最后1个
	// 但这样归并的次数和从前往后两两归并的次数是一样的？所以还是 上述的从前往后好了
	// 复杂度是 19*500
	Integer[] result2 = data[0];
	for (int i = 1; i < rowSize; i++) {
		result2 = merge(result2, data[i], columnSize);
	}

	System.out.println("method 2 result:        " + Arrays.toString(result2));

	System.out.println("result1 equals result2: " + Arrays.equals(result1, result2));

	// 方法3，保持一个最小堆，这个堆存放来自20个数组的最小数
	// 每次取出一个数，然后将该数所在的数组的后面一个数入堆
	// 重复上面步骤，取出500个数
	// 注意建堆的时候需要保持 数来自哪个数组，用一个内部类实现
	// 复杂度是 500 * log(20)
	Integer[] result3 = new Integer[500];
	MinHeap<DataWithSource> heap = new MinHeap<DataWithSource>();
	for (int i = 0; i < rowSize; i++) {
		// 记录下来源那个数组，以及在数组中的 index
		DataWithSource d = new DataWithSource(data[i][0], i, 0);
		heap.add(d);
	}

	int num = 0;
	while (num < columnSize) {
		// 删除顶点元素
		DataWithSource d = heap.removeTop();
		result3[num++] = d.getValue();

		// 将 value 置为该数原数组里的下一个数
		d.setValue(data[d.getComeFrom()][d.getIndex() + 1]);

		// 将其在数组中的 index +1
		d.setIndex(d.getIndex() + 1);
		heap.add(d);
	}
	
	System.out.println("method 3 result:        " + Arrays.toString(result3));

	System.out.println("result2 equals result3: " + Arrays.equals(result2, result3));


}

输出

method 1 result:        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 28, 28, 29, 29, 29, 29, 29, 29, 29, 29, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 32, 32, 32, 32, 32, 33, 33, 34, 34, 34, 34, 34, 34, 34, 35, 35, 35, 35, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 37, 37, 37, 37, 38, 38, 38, 38, 38, 38, 38, 39, 39, 39, 39, 39, 39, 39, 39, 40, 40, 40, 40, 41, 41, 41, 41, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 42, 43, 43, 43, 43, 44, 44, 44, 44, 44, 45, 45, 45, 45, 45, 45, 45, 45, 45, 46, 46, 46, 46, 46, 46, 47, 47, 48, 48, 48, 49, 49, 49, 50, 50, 50, 50, 50, 51, 51, 51, 51, 51, 51, 51, 51, 52, 52, 52, 52, 52, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 54, 54, 54, 54, 54, 54, 54, 54, 55, 55, 55, 55, 55, 55, 56, 56, 56, 56, 56, 56, 56, 57, 57, 57, 57, 57, 57, 57, 58, 58, 58, 58, 58, 58, 58, 58, 58, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 60, 60, 60, 60, 61, 61, 61, 61, 61, 61, 61, 61, 61, 62, 62, 62, 62, 62, 62, 62, 62, 63, 63, 63, 63, 63, 63, 63, 64, 64, 64, 65, 65, 65, 65, 65, 65, 65, 65, 65, 65, 66, 66, 66, 66, 66, 66, 66, 66, 66, 67, 67, 67, 68, 68, 68, 68, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69, 70, 70, 70, 70, 71, 71, 71, 71, 71, 71, 71, 72, 72, 72, 72, 72, 72, 72, 73, 73, 73, 73, 73, 74, 74, 74, 74, 74, 74, 74, 75, 75, 76, 76, 76, 76, 76, 76, 76, 76, 76, 77, 77, 77, 77, 77, 77, 78, 78, 78, 78, 78, 78, 79, 79, 79]
method 2 result:        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 28, 28, 29, 29, 29, 29, 29, 29, 29, 29, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 32, 32, 32, 32, 32, 33, 33, 34, 34, 34, 34, 34, 34, 34, 35, 35, 35, 35, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 37, 37, 37, 37, 38, 38, 38, 38, 38, 38, 38, 39, 39, 39, 39, 39, 39, 39, 39, 40, 40, 40, 40, 41, 41, 41, 41, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 42, 43, 43, 43, 43, 44, 44, 44, 44, 44, 45, 45, 45, 45, 45, 45, 45, 45, 45, 46, 46, 46, 46, 46, 46, 47, 47, 48, 48, 48, 49, 49, 49, 50, 50, 50, 50, 50, 51, 51, 51, 51, 51, 51, 51, 51, 52, 52, 52, 52, 52, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 54, 54, 54, 54, 54, 54, 54, 54, 55, 55, 55, 55, 55, 55, 56, 56, 56, 56, 56, 56, 56, 57, 57, 57, 57, 57, 57, 57, 58, 58, 58, 58, 58, 58, 58, 58, 58, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 60, 60, 60, 60, 61, 61, 61, 61, 61, 61, 61, 61, 61, 62, 62, 62, 62, 62, 62, 62, 62, 63, 63, 63, 63, 63, 63, 63, 64, 64, 64, 65, 65, 65, 65, 65, 65, 65, 65, 65, 65, 66, 66, 66, 66, 66, 66, 66, 66, 66, 67, 67, 67, 68, 68, 68, 68, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69, 70, 70, 70, 70, 71, 71, 71, 71, 71, 71, 71, 72, 72, 72, 72, 72, 72, 72, 73, 73, 73, 73, 73, 74, 74, 74, 74, 74, 74, 74, 75, 75, 76, 76, 76, 76, 76, 76, 76, 76, 76, 77, 77, 77, 77, 77, 77, 78, 78, 78, 78, 78, 78, 79, 79, 79]
result1 equals result2: true
method 3 result:        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 28, 28, 29, 29, 29, 29, 29, 29, 29, 29, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 32, 32, 32, 32, 32, 33, 33, 34, 34, 34, 34, 34, 34, 34, 35, 35, 35, 35, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 37, 37, 37, 37, 38, 38, 38, 38, 38, 38, 38, 39, 39, 39, 39, 39, 39, 39, 39, 40, 40, 40, 40, 41, 41, 41, 41, 41, 41, 41, 41, 42, 42, 42, 42, 42, 42, 42, 43, 43, 43, 43, 44, 44, 44, 44, 44, 45, 45, 45, 45, 45, 45, 45, 45, 45, 46, 46, 46, 46, 46, 46, 47, 47, 48, 48, 48, 49, 49, 49, 50, 50, 50, 50, 50, 51, 51, 51, 51, 51, 51, 51, 51, 52, 52, 52, 52, 52, 53, 53, 53, 53, 53, 53, 53, 53, 53, 53, 54, 54, 54, 54, 54, 54, 54, 54, 55, 55, 55, 55, 55, 55, 56, 56, 56, 56, 56, 56, 56, 57, 57, 57, 57, 57, 57, 57, 58, 58, 58, 58, 58, 58, 58, 58, 58, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 60, 60, 60, 60, 61, 61, 61, 61, 61, 61, 61, 61, 61, 62, 62, 62, 62, 62, 62, 62, 62, 63, 63, 63, 63, 63, 63, 63, 64, 64, 64, 65, 65, 65, 65, 65, 65, 65, 65, 65, 65, 66, 66, 66, 66, 66, 66, 66, 66, 66, 67, 67, 67, 68, 68, 68, 68, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69, 69, 70, 70, 70, 70, 71, 71, 71, 71, 71, 71, 71, 72, 72, 72, 72, 72, 72, 72, 73, 73, 73, 73, 73, 74, 74, 74, 74, 74, 74, 74, 75, 75, 76, 76, 76, 76, 76, 76, 76, 76, 76, 77, 77, 77, 77, 77, 77, 78, 78, 78, 78, 78, 78, 79, 79, 79]
result2 equals result3: true