归并排序
链接:https://www.nowcoder.com/questionTerminal/96bd6684e04a44eb80e6a68efc0ec6c5 来源:牛客网
public
class
Solution {
public
int
InversePairs(
int
[] array) {
if
(array==
null
||array.length==
0
)
{
return
0
;
}
int
[] copy =
new
int
[array.length];
for
(
int
i=
0
;i<array.length;i++)
{
copy[i] = array[i];
}
int
count = InversePairsCore(array,copy,
0
,array.length-
1
);
//数值过大求余
return
count;
}
private
int
InversePairsCore(
int
[] array,
int
[] copy,
int
low,
int
high)
{
if
(low==high)
{
return
0
;
}
int
mid = (low+high)>>
1
;
int
leftCount = InversePairsCore(array,copy,low,mid)%
1000000007
;
int
rightCount = InversePairsCore(array,copy,mid+
1
,high)%
1000000007
;
int
count =
0
;
int
i=mid;
int
j=high;
int
locCopy = high;
while
(i>=low&&j>mid)
{
if
(array[i]>array[j])
{
count += j-mid;
copy[locCopy--] = array[i--];
if
(count>=
1000000007
)
//数值过大求余
{
count%=
1000000007
;
}
}
else
{
copy[locCopy--] = array[j--];
}
}
for
(;i>=low;i--)
{
copy[locCopy--]=array[i];
}
for
(;j>mid;j--)
{
copy[locCopy--]=array[j];
}
for
(
int
s=low;s<=high;s++)
{
array[s] = copy[s];
}
return
(leftCount+rightCount+count)%
1000000007
;
}
}