题目

• 在数组中的两个数字，如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组,求出这个数组中的逆序对的总数P。并将P对1000000007取模的结果输出。 即输出P%1000000007

代码

• 归并排序的增强版本，根据左神的代码修改，以归并快排为基础

public class Solution {
    long count=0;
    public int InversePairs(int [] arr) {
        if (arr == null || arr.length < 2) {
			return 0;
		}
		mergeSort(arr, 0, arr.length - 1);
        return count;
    }
    public static void mergeSort(int[] arr, int l, int r) {
		if (l == r) {
			return;
		}
		int mid = l + ((r - l) >> 1);
		mergeSort(arr, l, mid);
		mergeSort(arr, mid + 1, r);
		merge(arr, l, mid, r);
	}

	public static void merge(int[] arr, int l, int m, int r) {
		int[] help = new int[r - l + 1];
		int i = 0;
		int p1 = l;
		int p2 = m + 1;
		while (p1 <= m && p2 <= r) {
            if(arr[p1]>arr[p2]){
                help[i++]=arr[p2++];
                count+=m-p1+1;
                count=count>1000000007?(int)(count%1000000007):count;
            }else
                help[i++]=arr[p1++];
		}
		while (p1 <= m) {
			help[i++] = arr[p1++];
		}
		while (p2 <= r) {
			help[i++] = arr[p2++];
		}
		for (i = 0; i < help.length; i++) {
			arr[l + i] = help[i];
		}
	}
}

• 牛客代码

public class Solution {
    int count=0;
    public int InversePairs(int [] array) {
        if(array==null||array.length<=0) return 0;
        mergeUp2Down(array,0,array.length-1);
        return count;
    }
    public void mergeUp2Down(int [] a,int start,int end){
        if(start>=end) return;
        int mid=(end+start)>>1;
        mergeUp2Down(a,start,mid);
        mergeUp2Down(a,mid+1,end);
        merge(a,start,mid,end);
    }
    public void merge(int [] a,int start,int mid,int end){
        int[] temp=new int[end-start+1];
        int i=start,j=mid+1,index=0;
        while(i<=mid&&j<=end){
            if(a[i]>a[j]) {
                temp[index++]=a[j++];
                count+=mid-i+1;
                count=count>1000000007?count%1000000007:count;
            }else 
                temp[index++]=a[i++];
        }
        while(i<=mid) 
            temp[index++]=a[i++];
        while(j<=end) 
            temp[index++]=a[j++];
        for(int k=0;k<temp.length;k++) 
            a[start+k]=temp[k];
    }
}