【hihocoder】三十九周：二分.归并排序之逆序对

就是用归并排序求数组中得逆序对。假设数组为a:[2 4 5],和b:[1 3]，那么在这一次归并的时候逆序对这样求,belement表示当前result数组中b数组对应的元素个数,total表示逆序对的个数：

a:[2 4 5] b:[1 3] result{}

a:[2 4 5] b[3] result{1} belement = 1;

a:[4 5] b[3] result{1 2} belement = 1; total = total + belement = 1;

a:[4 5] b[] 　 result{1 2 3} belement = 2; total = 1;

a:[5] b[] result{1 2 3 4} belement = 2; total = total + belement = 3

a:[] b[] result{1 2 3 4 5} belement = 2; total = total + belement = 5

所以数组2 4 5 1 3的逆序数总共有5个。

JAVA版本代码如下：注意total要设置成long型防止溢出。

import java.util.Scanner;
public class Main {

public static void main(String[] args) {
// TODO Auto-generated method stub

Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] arr = new int
;
for(int i = 0;i < n;i++)
arr[i] = in.nextInt();
MergeSort(arr,0,arr.length-1);
/*
for(int i = 0;i < arr.length;i++)
System.out.print(arr[i]);
*/
System.out.print(total);
}
private static void MergeSort(int[] arr,int begin,int end){
if(begin >= end)
return;
int mid = (begin+end)/2;
MergeSort(arr, begin, mid);
MergeSort(arr, mid+1, end);
merge(arr, begin, end);
}
public static long total = 0;
private static void merge(int[] arr,int begin,int end){
int belement = 0;
int mid = (begin+end)/2;
int p1 = begin;
int p2 = mid+1;
int count = 0;
int[] sorted = new int[end-begin+1];

while(p1 <= mid && p2 <= end){
if(arr[p1] > arr[p2]){
sorted[count] = arr[p2];
p2++;
count++;
belement++;
//System.out.println(belement);
}else{
total += belement;
sorted[count] = arr[p1];
p1++;
count++;
}
}

while(p1 <= mid){
sorted[count] = arr[p1];
count++;
p1 ++;
total += belement;
}

while(p2 <= end){
sorted[count] = arr[p2];
count ++;
p2++;
}

for(int i = begin;i <= end;i++)
arr[i] = sorted[i-begin];
}

}

View Code