算法导论　　思考题2-4逆序对问题：设A[1..n]是一个包含n个不同数的数组。如果在i<j的情况下，有A[i]>A[j]，则称( i , j )就称为A中的一个逆序对(inversion)。给出一个算法，它能用Θ(nlgn)的最坏情况运行时间，确定n个元素的任何排列中逆序对的数目。（提示：修改合并算法）-Algorithm is graded 2-4 reverse introduction to question: set A [1.. n] is A contains n A different number of array. If in the I < j, there may be A [I] > A [j], says (I, j) is called A inversion of (one of the reverse. Give an algorithm, it can NLGN) Θ (the worst running time, determine any arrangement-- a elements in the number of the reverse. (hint: modify merger algorithm)