一、思路
(1)确定分界点:mid=(l+r)/2 ------这里和快排不同
(2)递归排序(left right)
(3)归并------合二为一
时间复杂度nlogn
二、题目练习
三、模板
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int q[N],tmp[N];
int n;
void merge_sort(int q[],int l,int r)
{
if(l>=r)return;
int mid=l+r>>1;
merge_sort(q,l,mid),merge_sort(q,mid+1,r);
int k=0,i=l,j=mid+1;
while(i<=mid&&j<=r)
if(q[i]<=q[j])tmp[k++]=q[i++];
else tmp[k++]=q[j++];
while(i<=mid)tmp[k++]=q[i++];
while(j<=r)tmp[k++]=q[j++];
for(i=l,j=0;i<=r;i++,j++)q[i]=tmp[j];
}
int main()
{
scanf("%d",&n);
for(int i=0;i<n;i++)scanf("%d",&q[i]);
merge_sort(q,0,n-1);
for(int i=0;i<n;i++)printf("%d ",q[i]);
return 0;
}
#include<iostream>
using namespace std;
const int N = 1e5+10;
int q[N],tmp[N];
int n;
long long result=0;
void merge_sort(int q[],int l, int r)
{
if(l>=r)return;
int mid=l+r>>1;
merge_sort(q,l,mid),merge_sort(q,mid+1,r);
int k=0,i=l,j=mid+1;
while(i<=mid&&j<=r)
if(q[i]<=q[j])tmp[k++]=q[i++];
else
{
tmp[k++]=q[j++];
result+=mid-i+1;
}
while(i<=mid)tmp[k++]=q[i++];
while(j<=r)tmp[k++]=q[j++];
for(int i=l,j=0;i<=r;i++,j++)q[i]=tmp[j];
}
int main()
{
scanf("%d",&n);
for(int i=0;i<n;i++)scanf("%d",&q[i]);
merge_sort(q,0,n-1);
printf("%lld",result);
return 0;
}