1、编程实现随机输入一个长度为n的数组A[n],求出数组A[n]中逆序对的总数。要求用两种方法实现,并分析其时间复杂度:1)枚举法2)分治法
2、编程实现求数组中第K小的元素。
枚举法代码
public class InverseNumber {
public static void main(String[] args) {
int sum=0;
int a[]= {5,4,3,2,1};
for (int i = 0; i <a.length ; i++) {
for (int j = i+1; j < a.length; j++) {
if(a[i]>a[j]) {
sum++;
}
}
}
System.out.println(sum);
}
}
分治法代码:
import java.util.Scanner;
public class InverseNumber01 {
public static int mergeSort(int[] arr,int x,int y,int []temp){
if(y-x<=1)
return 0;
int mid=(x+y)/2;
int left=mergeSort(arr,x,mid,temp);
int right=mergeSort(arr,mid,y,temp);
int index=x,p=x,q=mid;
int count=0;
while(p<mid||q<y){
if(p>=mid){
temp[index++]=arr[q++];
}else if(q>=y){
temp[index++]=arr[p++];
}else if(arr[p]<=arr[q]){
temp[index++]=arr[p++];
}else{
temp[index++]=arr[q++];
count+=(mid-p);
}
}
for(int i=x;i<y;i++){
arr[i]=temp[i];
}
return count+left+right;
}
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext())
{
int n=scanner.nextInt();
int[] arr=new int[n];
int[] temp=new int[n];
for(int i=0;i<n;i++){
arr[i]=scanner.nextInt();
}
System.out.println(mergeSort(arr,0,n,temp));
}
scanner.close();
}
}