Title Descriptionin the array of two digits, if the previous number is greater than the number that follows, then these two numbers form an inverse pair. Enter an array to find the total number of reverse pairs in this array.
Thinking of solving problems
idea one: Violent search, a judgment of each number, the number after which there is no smaller than it, if there is a count plus one. Time O (n^2)
thinking Two: Using the merge sorting method, when merging, calculate the number of reverse order. See the code for the specific process. Time O (nlogn)
Implementation Code
Class Solution {Public:int inversepairs (vector<int> data) {if (Data.empty ()) return 0; int len = Data.size (); vector<int> copy (len,0); int result=0; result = MergeSort (data, copy,0,len-1); return result; } int MergeSort (vector<int> &data, vector<int> ©, int start, int end) {if (start = = E nd) {Copy[start] = Data[end]; return 0; } int mid = (End+start)/2; int left = MergeSort (Data,copy,start, mid); int right = MergeSort (data,copy, mid+1,end); int count = merge (Data,copy,start,mid,end); Return left + right +count; } int Merge (vector<int> &data, vector<int> ©, int start, int mid, int end) {int Lbeg in = start, LEnd = mid; int rbegin = mid+1, rEnd = end; int k=end; int count = 0; while (lEnd >= lbegin &&Amp REnd >= rbegin) {if (Data[lend] > Data[rend]) {count + = rend-rbegin+1;//calculated in reverse order copy[k--] = data[lend--]; } else copy[k--] = data[rend--]; } while (Lbegin <= lEnd) copy[k--] = data[lend--]; while (Rbegin <= rEnd) copy[k--] = data[rend--]; for (int i=start; i<=end; i++) data[i] = Copy[i]; return count; }};
Interview questions 36_ in the array of inverse pairs