Exercises-reverse pairs in an array

Reversed pairs in the array

1. Title description

• For two numbers in the array, if the first number is greater than the following number, the two numbers form a reverse pair. Enter an array to find the total number of reverse pairs in this array.
• Input and output description
  

2. Code implementation and ideas

The first type: violence law

public class TestDemo2 {
    
    
    public static int count(int[] A, int n) {
    
    
        if(n == 0)return 0;
        int count = 0;
        for(int i = 0;i < n;i++){
    
    
            for(int j = i;j < n;j++){
    
    
                if(A[i] > A[j]){
    
    
                    count++;
                }
            }
        }
        return count;
    }
    public static void main(String[] args) {
    
    
        int[] a = {
    
    1,2,3,4,5,6,7,0};
        int k = count(a,a.length);
        System.out.println(k);
    }
}

The second: use the idea of ​​merging and sorting

public class TestDemo2 {
    
    
    public static void main(String[] args) {
    
    
        int[] a = {
    
    1,2,3,4,5,6,7,0};
        int[] b = {
    
    7,5,6,4};
        int k = count(a,a.length);
        System.out.println(k);
        System.out.println(count(b,b.length));
    }
    public static int count(int[] A, int n) {
    
    
        if(n == 0)return 0;
        return mergesort(A,0,n-1);
    }
    //先分
    public static int mergesort(int[] arr,int low,int high){
    
    
        if(low == high)return 0;
        int mid = (low+high)/2;
        return mergesort(arr,low,mid)+mergesort(arr,mid+1,high)+merge(arr,low,high,mid);
    }
    public static int merge(int[] arr,int low,int high,int mid){
    
    
        int[] tmp = new int[high-low+1];//临时存放合并后的数组
        int k = 0;
        int count = 0;//统计次数
        int s1 = low;
        int s2 = mid+1;
        while(s1 <= mid && s2 <= high){
    
    
            if(arr[s1] <= arr[s2]){
    
    
                tmp[k++] = arr[s1++];
            }else {
    
    
                tmp[k++] = arr[s2++];
                count += (mid-s1+1);//因为当s1大于s2的时候，s1和mid之间的数字将全部大于s2，因为s1和mid之间是已经排好序的数组
            }
        }
        //有一个还没有装完
        while (s1 <= mid){
    
    
            tmp[k++] = arr[s1++];
        }
        while (s2 <= high){
    
    
            tmp[k++] = arr[s2++];
        }
        for(int i = 0;i < tmp.length;i++){
    
    
            arr[low+i] = tmp[i];
        }
        return count;
    }
}