求一个数组中的逆序对

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数组中的逆序对是指左边元素大于右边元素，这样的一对数就构成了一个逆序对。求一个数组中的逆序对。这里使用分治方法，具体的思路可以参考剑指offer中的一个问题，这里使用了归并的思想，假如我们将一个数组分成两个数组，而且这两个数组是有序的，在从尾部到头部合并的时候，就可以判定有多少对逆序的元素对

#include <iostream>#include <string>#include <vector>using namespace std;typedef long long llong;llong MergeAndCount(vector<unsigned int> &a, int left, int mid, int right, vector<unsigned int> &temp);llong MergeSortAndCount(vector<unsigned int>& a, int left, int right, vector<unsigned int>& temp);llong CountInversions(vector<unsigned int>& a);int main(){    //freopen("data.in", "r", stdin);    int N;    cin >> N;    vector<unsigned int> a(N, 0);    for (int i = 0; i < N; ++i)    {        cin >> a[i];    }    llong result = CountInversions(a);    cout << result << endl;    return 0;}llong MergeAndCount(vector<unsigned int> &a, int left, int mid, int right, vector<unsigned int> &temp){    int i = left;    int j = mid + 1;    int k = left;    llong cnt = 0;    while (i <= mid && j <= right)    {        if (a[i] <= a[j])        {            temp[k++] = a[i++];        }        else         {            temp[k++] = a[j++];            cnt += mid - i + 1;    // key step        }    }    while (i <= mid) temp[k++] = a[i++];    while (j <= right) temp[k++] = a[j++];    for (i = left; i <= right; ++i)    {        a[i] = temp[i];    }    return cnt;}llong MergeSortAndCount(vector<unsigned int>& a, int left, int right, vector<unsigned int>& temp){    // base case    if (left >= right) return 0;    int mid = (left + right) / 2;        llong InversionCnt1 = MergeSortAndCount(a, left, mid, temp);    llong InversionCnt2 = MergeSortAndCount(a, mid+1, right, temp);    llong MergeInversionCnt = MergeAndCount(a, left, mid, right, temp);    return InversionCnt1 + InversionCnt2 + MergeInversionCnt;}llong CountInversions(vector<unsigned int>& a){    int n = a.size();    vector<unsigned int> temp(a.begin(), a.end());    llong ans = MergeSortAndCount(a, 0, n-1, temp);    return ans;}

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