36-数组中的逆序对

用到了归并排序的思想，有关归并排序：

白话经典算法系列之五 归并排序的实现

如果前半部分的数组中一个数大于后半部分的某个数，因为子数组是排好序的，所以肯定大于后半部分某数之前的全部数，即j-mid，见代码中，这就是i对应的逆序数目。

class Solution {public:    long long  InversePairsCore(vector<int>& data, vector<int>& copy, int start, int end){if (start == end)return 0;long long mid = (start + end) >>1;long long  left = InversePairsCore( copy,data, start, mid);  //这里把copy和data换了一下位置，滚动数组的思想？long long  right = InversePairsCore( copy,data, mid + 1, end);long long  i = mid;long long  j = end;long long  indexcopy = end;long long  count = 0;while (i >= start && j > mid){if (data[i] > data[j]){copy[indexcopy--] = data[i--];    count += j - mid;                      //见上方黑粗字}elsecopy[indexcopy--] = data[j--];}while (i >= start)copy[indexcopy--] = data[i--];while (j > mid)copy[indexcopy--] = data[j--];    //for(int s=start;s<=end;s++)     //   data[s]=copy[s];return left + right + count;}int InversePairs(vector<int> data){auto t = data.size();if (t == 0)return 0;vector<int> copy(data);long long count = InversePairsCore(data, copy, 0, t-1);copy.clear();return count % 1000000007;}};

自己写个归并排序：

void Merge(int data[], int copy[], int start, int end,int mid){int i = start;int  j = mid+1;int k = 0;while (i <= mid && j <=end){if (data[i] < data[j]){copy[k++] = data[i++];}elsecopy[k++] = data[j++];}while (i <= mid)copy[k++] = data[i++];while (j <= end)copy[k++] = data[j++];for (int i = 0; i < k;i++){data[start + i] = copy[i];}}void mergesort(int data[], int copy[], int start, int end){if (start < end){int mid = (start + end) >> 1;mergesort(data, copy, start, mid);mergesort(data, copy, mid + 1, end);Merge(data, copy, start, end,mid);}}bool mergesort(int a[],int n){int *p = new int[n];if (p == NULL)return false;mergesort(a,p, 0, n - 1);delete[] p;return true;}