(归并排序) 用swift的代码形式实现 首先是第一部分,这个function的目的是要把一个array分割成一个个小部分,这里值得注意的是有一个guard的关键字,我们想一把每一个array里的元素都分开来,直到array的count变成0,这里我return先只给一个空值,等会会有完成的code 这是第二个function,Merge sort的主要逻辑是把一个...
functioncountInversions($arr){$temp=array_fill(0,count($arr),0);returnmergeSort($arr,$temp,0,count($arr)-1);}functionmergeSort(&$arr,&$temp,$left,$right){$inversions=0;if($left<$right){$mid=(int)(($left+$right)/2);$inversions+=mergeSort($arr,$temp,$left,$mid);$inversions+...
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 ...
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...
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...
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 ...