def countInversions(arr): def merge_sort_and_count(arr): if len(arr) <= 1: return arr, 0 mid = len(arr) // 2 left, left_inversions = merge_sort_and_count(arr[:mid]) right, right_inversions = merge_sort_and_count(arr[mid:]) merged, split_inversions = merge_and_count(left, ...
Forthe second problem,I counted numberofnumbers greater than A[i]at its left(left[i])andnumberofnumbers less than A[i]at its right(right[i]).Sothe outputissummationofright[i]*left[i]forall ifrom0to n-1. Now I have a question, what if the problem is counting inversions of length ...
We sort the following datasets: is already sorted, so there are no inversions for us to correct. Thus, we print on a new line. We performed a total of swaps to correct inversions. 题目解析 核心就是Merge排序,注意在新建数组的时候直接赋值,append等会导致超时。就是用递归 每次都拿两个排序好的...
Solution: importjava.io.*;importjava.util.*;importjava.text.*;importjava.math.*;importjava.util.regex.*;publicclassSolution{publicstaticlongcountInversions(int[]arr){returnmergeSort(arr,0,arr.length-1);}publicstaticlongmergeSort(int[]arr,intstart,intend){if(start==end)return0;longcount=0;in...
The way in which we'll divide will be motivated directly by merge sort where we recurs e separately on the left and the right half's of the array. We're gonna do the same thing here. To understand how much progress we can make purely using recursion let's classify the inversions of ...