Give an O(lg n)-time algorithm to find the median of all 2n elements in arrays X and Y._find the median of all 2n numbers in o(log n) time-优快云博客

本文链接：https://blog.youkuaiyun.com/chmodexplorer/article/details/10400153

本文介绍了一种高效算法，可在两个已排序的数组中查找中位数，时间复杂度为O(log n)。该算法通过递归划分并比较元素来逐步逼近中位数，适用于需要快速处理大量数据的应用场景。

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Let X[1 ‥ n] and Y[1 ‥ n] be two arrays, each containing n numbers already in sorted order. Give anO(lg n)-time algorithm to find the median of all 2n elements in arraysX and Y.

    template <typename Iterator>
    std::pair<Iterator, Iterator> find_median( Iterator b1, Iterator e1, Iterator b2, Iterator e2 )
    {
        size_t n1 = std::distance( b1, e1 ), n2 = std::distance( b2, e2 );
        assert( n1 > 0 && n2 > 0 && n1 == n2 );
        if ( n1 > 2 ) {
            while ( n1 != 2 && n2 != 2 ) {
                auto m1 = b1 + n1/2, m2 = b2 + n2/2;
                if ( *m1 <= *m2 ) {
                    b1 = m1;
                    e2 = m2 + 1;
                    n1 = std::distance( b1, e1 );
                    n2 = std::distance( b2, e2 );
                    if ( n1 != n2 ) {
                        assert( ( n1 + 1 ) == n2 );
                        --b1;
                        ++n1;
                    }
                } else {
                    b2 = m2;
                    e1 = m1 + 1;
                    n1 = std::distance( b1, e1 );
                    n2 = std::distance( b2, e2 );
                    if ( n1 != n2 ) {
                        assert( ( n2 + 1 ) == n1 );
                        --b2;
                        ++n2;
                    }
                }
            }
            assert( n1 == n2 );
        }
        
        std::pair<Iterator, Iterator> medians;
        if ( n1 == 2 ) {
            if ( *b1 <= *b2 ) {
                if ( *b2 <= *( b1 + 1 ) ) {
                    medians.first = b2;
                    if ( *( b1 + 1 ) <= *( b2 + 1 ) ) {
                        medians.second = b1 + 1;
                    } else {
                        medians.second = b2 + 1;
                    }
                } else {
                    medians.first = b1 + 1;
                    medians.second = b2;
                }
            } else {
                if ( *b1 <= *( b2 + 1 ) ) {
                    medians.first = b1;
                    if ( *( b2 + 1 ) <= *( b1 + 1 ) ) {
                        medians.second = b2 + 1;
                    } else {
                        medians.second = b1 + 1;
                    }
                } else {
                    medians.first = b2 + 1;
                    medians.second = b1;
                }
            }
        } else {
            assert( n1 == 1 );
            if ( *b1 <= *b2 ) {
                medians.first = b1;
                medians.second = b2;
            } else {
                medians.first = b2;
                medians.second = b1;
            }
        }
        return medians;
    }

The testing code:

        for ( int retry = 0; retry < 64; ++retry ) {
            size_t size = get_random_size();
            std::vector<int> ai1( size );
            random_data( &ai1[ 0 ], sizeof( ai1[ 0 ] ) * size );
            std::vector<int> ai2( size );
            random_data( &ai2[ 0 ], sizeof( ai2[ 0 ] ) * size );

            std::sort( ai1.begin(), ai1.end() );
            std::sort( ai2.begin(), ai2.end() );

            auto mi = find_median( ai1.begin(), ai1.end(), ai2.begin(), ai2.end() );
            std::vector<int> buf( size * 2 );
            std::merge( ai1.begin(), ai1.end(), ai2.begin(), ai2.end(), buf.begin() );
            if ( *mi.first != buf[ size - 1 ] && *mi.second != buf[ size ] ) {
                //save_data( "d1.bin", &ai1[ 0 ], size );
                //save_data( "d2.bin", &ai2[ 0 ], size );
                auto i = std::lower_bound( buf.begin(), buf.end(), *mi.first );
                fprintf( stderr, "Cannot find the median correctly in the two sorted %u-size arrays, the false one was %u\n", size, i - buf.begin() );
                DebugBreak();
            }
        }