Reduce inversion count 求最小逆序数

本问题出自：微软2014实习生及秋令营技术类职位在线测试 (Microsoft Online Test for Core Technical Positions)

Description

Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array. If such a pair exists, return the new inversion count; otherwise returns the original inversion count.

在一个数列中选择两个数，交换他们的顺序使得逆序数变得最小，给出这个逆序数值。

Definition of Inversion: Let (A[0], A[1] ... A[n], n <= 50) be a sequence of n numbers. If i < j and A[i] > A[j], then the pair (i, j) is called inversion of A.

定义：在数列A[n]中，如果i < j 且A[i] > A[j]，那么(i, j)就是A的一个逆序。

Example:
Count(Inversion({3, 1, 2})) = Count({3, 1}, {3, 2}) = 2
InversionCountOfSwap({3, 1, 2})=>
{
 InversionCount({1, 3, 2}) = 1 <-- swapping 1 with 3, decreases inversion count by 1
 InversionCount({2, 1, 3}) = 1 <-- swapping 2 with 3, decreases inversion count by 1
 InversionCount({3, 2, 1}) = 3 <-- swapping 1 with 2 , increases inversion count by 1
}

Input

Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma.

Output

For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced.

Sample Input

3,1,2
1,2,3,4,5

Sample Output

1
0

解题思路

对于一个数列，首先求出其逆序数。

如果逆序数不等于0（不是顺序排列），那么一定有可以减小的方法（在一个排列中对换相邻的元素，如果把小的换到大的之前，那么逆序数减一）

一种做法是：

遍历所有的数对，对比这两个数和他们之间的数，由“在一个排列中对换相邻的元素，如果把小的换到大的之前，那么逆序数减一，反之加一”计算出减少的逆序数，选择减少最多的。