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

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://blog.csdn.net/never_cxb/article/details/50210317

方法思路分析

笔者想到3中方法,水平有限,欢迎拍砖指导,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(注意,我们不是归并得到100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. 网上的方法,就是堆。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
第2小的是将刚才的1替换为后面的元素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

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));

  
  
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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;

    }
  
  
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下来是总的归并代码

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    // 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();
    }

}

  
  
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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);
    }
}
  
  
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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));


    }

  
  
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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

  
  
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转载自blog.csdn.net/vipp666/article/details/78289356
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