Finding the median of an unsorted array

问题

To find the median of an unsorted array, we can make a min-heap in O(nlogn) time for n elements, and then we can extract one by one n/2 elements to get the median. But this approach would take O(nlogn) time.

Can we do the same by some method in O(n) time? If we can, then please tell or suggest some method.

回答1:

You can use the Median of Medians algorithm to find median of an unsorted array in linear time.

回答2:

I have already upvoted the @dasblinkenlight answer since the Median of Medians algorithm in fact solves this problem in O(n) time. I only want to add that this problem could be solved in O(n) time by using heaps also. Building a heap could be done in O(n) time by using the bottom-up. Take a look to the following article for a detailed explanation Heap sort

Supposing that your array has N elements, you have to build two heaps: A MaxHeap that contains the first N/2 elements (or (N/2)+1 if N is odd) and a MinHeap that contains the remaining elements. If N is odd then your median is the maximum element of MaxHeap (O(1) by getting the max). If N is even, then your median is (MaxHeap.max()+MinHeap.min())/2 this takes O(1) also. Thus, the real cost of the whole operation is the heaps building operation which is O(n).

BTW this MaxHeap/MinHeap algorithm works also when you don't know the number of the array elements beforehand (if you have to resolve the same problem for a stream of integers for e.g). You can see more details about how to resolve this problem in the following article Median Of integer streams

回答3:

Quickselect works in O(n), this is also used in the partition step of Quicksort.

回答4:

The quick select algorithm can find the k-th smallest element of an array in linear (O(n)) running time. Here is an implementation in python:

import random

def partition(L, v):
    smaller = []
    bigger = []
    for val in L:
        if val < v: smaller += [val]
        if val > v: bigger += [val]
    return (smaller, [v], bigger)

def top_k(L, k):
    v = L[random.randrange(len(L))]
    (left, middle, right) = partition(L, v)
    # middle used below (in place of [v]) for clarity
    if len(left) == k:   return left
    if len(left)+1 == k: return left + middle
    if len(left) > k:    return top_k(left, k)
    return left + middle + top_k(right, k - len(left) - len(middle))

def median(L):
    n = len(L)
    l = top_k(L, n / 2 + 1)
    return max(l)

回答5:

The answer is "No, one can't find the median of an arbitrary, unsorted dataset in linear time". The best one can do as a general rule (as far as I know) is Median of Medians (to get a decent start), followed by Quickselect. Ref: [https://en.wikipedia.org/wiki/Median_of_medians][1]

回答6:

It can be done using Quickselect Algorithm in O(n), do refer to Kth order statistics (randomized algorithms).

回答7:

As wikipedia says, Median-of-Medians is theoretically o(N), but it is not used in practice because the overhead of finding "good" pivots makes it too slow.
http://en.wikipedia.org/wiki/Selection_algorithm

Here is Java source for a Quickselect algorithm to find the k'th element in an array:

/**
 * Returns position of k'th largest element of sub-list.
 * 
 * @param list list to search, whose sub-list may be shuffled before
 *            returning
 * @param lo first element of sub-list in list
 * @param hi just after last element of sub-list in list
 * @param k
 * @return position of k'th largest element of (possibly shuffled) sub-list.
 */
static int select(double[] list, int lo, int hi, int k) {
    int n = hi - lo;
    if (n < 2)
        return lo;

    double pivot = list[lo + (k * 7919) % n]; // Pick a random pivot

    // Triage list to [<pivot][=pivot][>pivot]
    int nLess = 0, nSame = 0, nMore = 0;
    int lo3 = lo;
    int hi3 = hi;
    while (lo3 < hi3) {
        double e = list[lo3];
        int cmp = compare(e, pivot);
        if (cmp < 0) {
            nLess++;
            lo3++;
        } else if (cmp > 0) {
            swap(list, lo3, --hi3);
            if (nSame > 0)
                swap(list, hi3, hi3 + nSame);
            nMore++;
        } else {
            nSame++;
            swap(list, lo3, --hi3);
        }
    }
    assert (nSame > 0);
    assert (nLess + nSame + nMore == n);
    assert (list[lo + nLess] == pivot);
    assert (list[hi - nMore - 1] == pivot);
    if (k >= n - nMore)
        return select(list, hi - nMore, hi, k - nLess - nSame);
    else if (k < nLess)
        return select(list, lo, lo + nLess, k);
    return lo + k;
}

I have not included the source of the compare and swap methods, so it's easy to change the code to work with Object[] instead of double[].

In practice, you can expect the above code to be o(N).

来源：https://stackoverflow.com/questions/10662013/finding-the-median-of-an-unsorted-array

标签

algorithm

heap

median