Wednesday, March 2, 2016

问题 - 求矩阵中第K大的数是多少? - 51Nod


问题 - 求矩阵中第K大的数是多少? - 51Nod
2个整数数组A,B,长度分别为m,n。从A,B中各选1个元素A[i],B[j],相加后得到C[i,j],共有m*n种选择方式,对应m*n个数,求这m*n个数中第K大的数是多少?


1、碰到这个问题的一般思路是,把m*n个数先求出来,然后用Nth-Elemnet,找出第k大的数。Nth-Element就是快排变形的那个partition,线性时间,这里不延伸讲了。这样的话时间复杂度O(m*n),空间复杂度O(m*n)。

2、对于第一种方法,时间复杂度还好说,空间复杂度实在问题比较大,是否可以不求出这m*n个元素就直接找打第K大的数呢?

可以这样,先对A,B分别排序。复杂度n*log(n) + m*log(m),排序后我们就发现第k大的数一定在Min = A[0] + B[0] <= C[k] <= A[n] + B[m] = Max,而且从Min到Max,K是递减,如果用二分法找这个C[k],然后在A和B中判断,有多少个数比C[k]大。如果恰好有k-1个数比C[k]大,则C[k]是第K大的数。考虑有相等的情况,如果比C[K]大的数不到K个,而比C[K]小的数不到m*n - k个,则说明等于C[K]的数有多个(当然直接判断=也可以),那么C[K]就是第k大的数。

总结一下这个思路,就是在Min和Max之间,二分找这个和,然后在排序好的数组A,B中,统计有多少个数大于这个和。

在这个思路的基础上,如何统计有多少个数的和大于给定的数C[k]呢?我们可以枚举数组A中的元素,然后利用二分,在数组B中找到C[k] - A[i]对应的元素B[j],那么所有比B[j]大的元素同A[i]的和,都大于C[k]。这样就可以写出一个n*log(m)*log(Max - Min) + n*log(n) + m*log(m)的程序来解决这个问题,空间复杂度O(1)

3、对于方法2,有一个简单的优化,就是如果n > m,也就是数组B的长度大于数组A的长度,如果枚举数组B,效率会更高,空间还是O(1),时间变为了min(m,n)*log(max(n,m))*log(max - min) + n*log(n) + m*log(m)

4、对于方法2和3,都还可以进一步优化,优化的关键点在于:枚举数组A,在数组B中二分找一个数。如果改用头尾指针的方式,平摊下来是线性的。方法如下:

先用一个指针指向数组A的头,一个指向数组B的尾,如果A[0] + B[m] < C[k],A的指针后移,判断A[1] + B[m],如果A[0] + B[m] > C[k],则B的指针前移,判断A[0] + B[m-1]。简单说就是< C[k],A的指针后移,> C[k],B的指针前移。不过问题的关键在于计数,同时统计大于C[k]的小于C[k]的数有多少个。如果A[0] + B[m] < C[k],则A[0] + B[0],B[1]......B[m-1]都小于<C[k],小于的计数+m,如果A[0] + B[m] > C[k],则A[1],A[2]......A[n]+B[m]都大于C[k],大于计数+n,这样,就可以写出一个(m+n)*log(Max-Min)+ n*log(n) + m*log(m)的程序,来解决这个问题。

到此为止,优化到头了么?其实此问题可以转为在杨氏矩阵中找第k大的值,问题还有优化的空间,关键在于Log(Max - Min),可以优化到Log(m*n),后者适用范围就不仅仅是整数了,实数范围都可以,但具体方法比较复杂,有兴趣大家可以再深入研究一下。
        static private int[] A, B;
        //求第K个元素

        static long KthElement(long k)
        {
            //该数肯定位于upper和lower之间
            long upper = (long)A[A.Length - 1] * B[B.Length - 1], lower = A[0] * B[0];
            return Solve(upper, lower, k);
        }

        //二分查找该值
        static long Solve(long upper, long lower, long count)
        {
            if (upper == lower) return upper;

            long mid = (upper + lower) >> 1;

            if (Counter(mid) >= count)
                return Solve(upper, mid + 1, count);

            if (Counter(mid - 1) < count)
                return Solve(mid - 1, lower, count);

            return mid;
        }

        //统计比指定value大的数有多少个
        static long Counter(long value)
        {
            int i = 0, j = B.Length - 1;
            long count = 0;

            while (i < A.Length && j >= 0)
            {
                if ((long)A[i] * (long)B[j] > value)
                {
                    count += A.Length - i;
                    j--;
                }
                else
                    i++;
            }
            return count;
        }
    }
http://blog.csdn.net/huangxy10/article/details/8084652
https://segmentfault.com/q/1010000000715010
http://zhiqiang.org/blog/science/computer-science/median-algorithm-of-ordered-matrix.html
给定 n×n 的实数矩阵,每行和每列都是递增的,求这 n2 个数的中位数。
使用类似Tarjan的线性中位数的方法,每次找每列中位数,然后找中位数的中位数,之后可以删除前一半列的上半部分或者后一半列的下半部分,这样可以实现复杂性 O(nlog2n) 。
但是这个问题是有 O(n) 的算法的,在Top Language Google GroupObtuse Sword的指点下,找到了这个问题的原始论文:Generalized Selection and Ranking: Sorted Matrices。事实上这篇论文证明了更强的结论:
对于一个 n×m ( nm )的矩阵,若每行和每列都是递增的,则可以在 O(nlog2m/n) 找到第 k大的数。
算法的基本思路是将矩阵依次对半划分成更小的子矩阵,然后删除不可能包含所求中位数的子矩阵。通过对每次划分后子矩阵个数的估计,发现此算法时间复杂度为 O(nlog2m/n) 。
使用同样的技巧,可以证明更更强的结论(这个算法具体过程就没细看了):
一堆 ni×mi ( nimi )的矩阵,若每个矩阵的每行和每列都是递增的,则selection problem(即找第 k 大的数)的时间复杂度为 O(nilog2mi/ni) 。
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