求逆序对的个数

特点：利用归并排序中合并的步骤，计算逆序对
时间复杂度O（nlgn）

int merge_inversion(int *arr,int start,int end,int middle);
int count_inversion(int *arr,int start,int end)
{if(start<end){int middle = (start + end)/2;int left = count_inversion(arr,start,middle);int right = count_inversion(arr,middle+1,end);return left + right + merge_inversion(arr,start,end,middle);}elsereturn 0;
}
int merge_inversion(int *arr,int start,int end,int middle)
{int *left_arr = new int [middle - start + 2];int *right_arr = new int [end - middle + 3];for (int i = start; i < middle+1; ++i) {left_arr[i-start] = arr[i];}for (int i = middle+1; i < end+1; ++i) {right_arr[i-middle-1] = arr[i];}left_arr[middle -start + 1] = INT_MAX;right_arr[end - middle] = INT_MAX;int left = 0,right = 0,inversions = 0;int left_count = middle - start + 1;for (int i = start; i < end+1; ++i) {if(left_arr[left]<=right_arr[right]){arr[i] = left_arr[left];left++;}else{arr[i] = right_arr[right];right++;inversions += left_count - left;}}delete [] left_arr;delete [] right_arr;return inversions;
}