出一个小题目:求数组的逆序数
即求数组有多少个 使得a在b前面,但是 a>b
例子:
[1 3 2] 的逆序数是1(即 <3,2> 一个)
[3 2 1] 的逆序数是3(即 <3,2> <2,1> <3,1> 三个)
3P版:
求数组有多少个 使得a在b前面,b在c前面,但是 a>b>c
1.暴力解法
#include <stdio.h>
int fun_1(int *base,int x,int y);
int main(void)
{
int A[]={8,7,6,5,10};
printf("%d",fun_1(A,0,4));
return 0;
}
int fun_1(int *base,int x,int y) //插入排序修改 冒泡排序类似 元素交换次数即为所求
{
int num=0;
for(int i=x+1;i<=y;i++)
{
for(int j=i-1;j>=0;j--)
{
if(base[j]>base[i])
num++;
}
}
return num;
}
2.归并排序
运用分冶法的思想
#include <stdio.h>
void merge(int *base,int x,int mid,int y,int *tmp);
void msort(int *base,int x,int y,int *tmp);
int num=0;
int main(void)
{
int A[10]={10,9,8,7,6,5,4,3,2,1};
int tmp[10];
msort(A,0,9,tmp);
printf("%d",num);
return 0;
}
void msort(int *base,int x,int y,int *tmp)
{
if(x<y)
{
int mid=x+(y-x)/2;
msort(base,x,mid,tmp);
msort(base,mid+1,y,tmp);
merge(base,x,mid,y,tmp);
}
}
void merge(int *base,int first,int mid,int last,int *temp)
{
int i=first;
int j=mid+1;
int k=0;
while(i<=mid&&j<=last)
{
if(base[i]<base[j])
temp[k++]=base[i++];
else
{
temp[k++]=base[j++];
num=num+mid-i+1;
}
}
while(i<=mid)
temp[k++]=base[i++];
while(j<=last)
temp[k++]=base[j++];
for(int v = 0; v < k; v++)
base[first + v] = temp[v];
}
3.运用树状数组
#include <stdio.h>
int lowbit(int x) //定义一个Lowbit函数,返回参数转为二进制后,最后一个1的位置所代表的数值.
{
return x&(-x);
}
void insert(int i,int x,int *c)
{
while(i<=10)
{
c[i]+=x;
i+=lowbit(i);
}
}
int getsum(int i,int *c)
{
int sum=0;
while(i>0)
{
sum+=c[i];
i-=lowbit(i);
}
return sum;
}
int main(void)
{
int A[11]={0,10,9,8,7,6,5,4,3,2,1};//A[0]位置没用
int c[100]={0};
int res=0;
for(int i=1;i<=10;i++)
{
insert(A[i],1,c);
res+=i-getsum(A[i],c);
}
printf("%d",res);
return 0;
}
4.构建特殊二叉搜索树
#include <stdio.h>
#include <stdlib.h>
typedef struct trnode
{
int data;
int size; //子节点数量
struct trnode *parent; //父节点
struct trnode *left;
struct trnode *right;
} Trnode;
Trnode* insert(int data,Trnode *head);
Trnode* MakeNode(int data);
void AddNode(Trnode*q,Trnode *head);
int main(void)
{
int A[10]={10,9,8,7,6,5,4,3,2,1};
int sum=0,num=0;
Trnode *head=NULL;
head=insert(A[0],head);
for(int i=1;i<10;i++)
{
Trnode *target=insert(A[i],head);
num=0;
Trnode *p=target;
while(p->parent!=NULL)
{
if(p->parent->right==p)
num+=p->parent->size-p->size;
}
p=p->parent;
}
sum+=head->size-num;
}
printf("%d",sum);
return 0;
}
Trnode* MakeNode(int data)
{
Trnode* node=(Trnode*)malloc(sizeof(Trnode));
node->data=data;
node->parent=NULL;
node->size=0;
node->left=NULL;
node->right=NULL;
return node;
}
Trnode *insert(int data,Trnode *head)
{
if(head==NULL)
{
return MakeNode(data);
}
else
{
Trnode*q=MakeNode(data);
Trnode *p=head;
AddNode(q,p);
return q;
}
}
void AddNode(Trnode*q,Trnode *head)
{
if(q->data<head->data&&head->left!=NULL)
{
head->size++;
head->left->parent=head;
head=head->left;
AddNode(q,head);
}
else if(q->data>head->data&&head->right!=NULL)
{
head->size++;
head->right->parent=head;
head=head->right;
AddNode(q,head);
}
if(head->left==NULL&&q->data<head->data)
{
q->parent=head;
head->size++;
head->left=q;
return;
}
else if(head->right==NULL&&q->data>head->data)
{
q->parent=head;
head->size++;
head->right=q;
return;
}
}
Note:
树状数组:
http://www.cnblogs.com/ECJTUACM-873284962/p/6380245.html
http://www.cnblogs.com/hsd-/p/6139376.html
https://zh.wikipedia.org/wiki/%E6%A0%91%E7%8A%B6%E6%95%B0%E7%BB%84
http://www.hawstein.com/posts/binary-indexed-trees.html
http://poj.org/summerschool/1_interval_tree.pdf