inverse avi 分享 .net div hit tex ack delet

int InversePairs(int* data, int length)
{
if (data == NULL || length < 0)
return 0;
int *copy = new int[length];
for (int i = 0; i < length; ++i)
copy[i] = data[i];
int count = InversePairsCore(data, copy, 0, length - 1);
delete[] copy;
return count;
}
int InversePairsCore(int* data, int* copy, int start, int end)
{

if (start == end)
{
copy[start] = data[start];
return 0;
}
int length = (end - start) / 2;
int left = InversePairsCore(copy, data, start, start + length);
int right = InversePairsCore(copy, data, start + length + 1, end);
//i初始化為前半段最後一個數字的下標
int i = start + length;
//j初始化為後半段最後一個數字的下標
int j = end;
int indexCopy = end;
int count = 0;
while (i >= start&& j >= start + length + 1)
{
if (data[i] > data[j])
{
copy[indexCopy--] = data[i--];
count += j - start - length;
}
else
{
copy[indexCopy--] = data[j--];
}
}
for (; i >= start; --i)
copy[indexCopy--] = data[i];
for (; j >= start + length + 1; --j)
copy[indexCopy--] = data[j];
return left + right + count;
}

36.數組中的逆序對