解题思路
1.归并排序 – 分治
确定分界点;
递归排序;
归并合二为一;
2.快速排序 – 分治
确定分界点;
划分调整区间;
递归处理左右两段;
样例
acwing 788 逆序对的数量
算法1 – 归并排序
C++ 代码
#include <iostream>
using namespace std;
const int N = 1e6 + 10;
int n;
int q[N],tmp[N];
void merge_sort(int q[], int l, int r)
{
if (l == r) return;
int mid = (l + r) / 2;
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;
}
算法2 – 快速排序
C++ 代码
#include <iostream>
using namespace std;
const int N = 1e6 + 10;
int n;
int q[N];
void quick_sort(int q[], int l, int r)
{
if(l == r) return;
int i = l - 1, j = r + 1;
int x = q[(l + r) / 2];
while(i < j)
{
do i ++ ; while(q[i] < x);
do j -- ; while(q[j] > x);
if(i < j) swap(q[i], q[j]);
}
quick_sort(q, l, j);
quick_sort(q, j + 1, r);
}
int main()
{
scanf("%d", &n);
for(int i = 0; i < n; i ++ ) scanf("%d", &q[i]);
quick_sort(q, 0, n-1);
for(int i = 0; i < n; i ++ ) printf("%d ", q[i]);
return 0;
}