算法導論在第二章的最後2-4小題中提到了查找逆序對的問題。
其實我認爲,查找逆序對其實就是排序的過程,比如3,2,1
插入排序的過程就是3,2-3,1-2-3.需要3次交換位置,這正好也是逆序對的數目(這裏針對的只能是數組,如果是鏈表則不是這樣)。
所以他要求在nLgn的時間計算出逆序對的數量,正好可以採用歸併排序。
#include<stdio.h>
#include<stdlib.h>
int merge(int *A, int left ,int mid, int right)
{
int i;
int j;
int Tl[8];
int Tr[8];
int lnum;
int rnum;
int inversion = 0;
lnum = mid-left+1;
rnum = right - mid;
//assinment
for(i =0;i<lnum;i++)
Tl[i]=A[left+i];
for(j =0;j<rnum;j++)
Tr[j] = A[mid+j+1];
i = 0;
j =0;
while(i<(lnum)&&j<(rnum))
{
if(Tl[i]<Tr[j])
A[left++] = Tl[i++];
else
{
A[left++] = Tr[j++];
//for (4,5)(2,3)
//Tl[0]>Tr[0],so inversion + lnum;
//then Tl[1]>Tr[0] ,also +lnum
inversion+=lnum -i;
}
}
while(i<(lnum))
{
A[left++] = Tl[i++];
}
while(j<(rnum))
{
A[left++] = Tr[j++];
}
return inversion;
}
int merge_sort(int *A,int left,int right)
{
int mid;
int inversion = 0;
if(left<right)
{
mid = (left+right)/2;
//count steps
inversion += merge_sort(A,left,mid);
inversion += merge_sort(A,mid+1,right);
inversion += merge(A,left,mid,right);
}
return inversion;
}
int main()
{
int A[8] = {5,4,3,2,1,0,7,6};
int sum;
//merge sort
sum = merge_sort(A,0,7);
}