数组逆序对数

package Algorithm.compet;


public class ReversePairs {
    // 一定要实例化才能实例化其中的变量（静态变量）
    static int number;
    static int sum;
    static int mergeSort(int a[],int l,int r){

        if(l==r){
            return 0;
        }

        int m=(l+r)/2;
        mergeSort(a, m+1, r);
        mergeSort(a, l, m);
        sum+=merge(a,l,m,r);
        return sum;
    }

    static int merge(int a[],int l,int m,int r){
        //若想引用数组变量，面向对象则为 int a[]=new int[len];
        int save[]=new int[a.length];//最大数值扩容
        int mk=m+1;
        int ll=l;
        int cIndex=l;
        int count=0;
        while(l<=m&&mk<=r){
            if(a[l]<=a[mk]) 
                save[ll++]=a[l++];
            else{
                count+=(m-l+1);
                save[ll++]=a[mk++];
            }

        }
        //归并左边剩余的数
        while(l<=m){
        save[ll++]=a[l++];  
        }
        //归并右边剩余的数
        while(mk<=r){
            save[ll++]=a[mk++];
        }

/*      if(cIndex==0&&r==a.length-1){
            for (int i : save) {
                System.out.print(i+" ");
            }
        }*/

        System.out.println("第"+(++number)+"趟排序:\t");
        //从临时数组拷贝到原数组
         while(cIndex<=r){
                a[cIndex]=save[cIndex];
                //输出中间归并排序结果
                System.out.print(a[cIndex]+"\t");
                cIndex++;
            }
         System.out.println();
         return count;

    }
    public static void main(String[] args) {
        int a[]= new int[]{4,5,3,2};
        System.out.println("数组逆序对的个数是："+mergeSort(a, 0, a.length-1));
    }
}

result:5{分别是：(4,3)(4,2)(5,3)(5,2)(3,2)}