題目描述
在數組中的兩個數字，如果前面一個數字大于后面的數字，則這兩個數字組成一個逆序對。輸入一個數組,求出這個數組中的逆序對的總數P。并將P對1000000007取模的結果輸出。 即輸出P%1000000007

public class Solution {
    
    int sum = 0;
    public int InversePairs(int[] array) {
        
        if(array == null || array.length == 0)
            return 0;
        mSort(array, 0, array.length - 1);
        return sum%1000000007;
    }
    private void mSort(int[] a, int left, int right) {
        
        if(left >= right)
            return;
        int mid = (left + right) / 2;
        mSort(a, left, mid);
        mSort(a, mid + 1, right);
        merge(a, left, mid, right);
    }
    private void merge(int[] a, int left, int mid, int right) {
        
        int i = mid;
        int j = right;
        int[] temp = new int[right - left + 1];
        int k = temp.length - 1;
        while(i >= left && j > mid) {
            
            if(a[i] <= a[j]) {
                
                temp[k] = a[j];
                j --;
                k --;
            }else {
                
                sum += (j - mid);
                sum %= 1000000007;
                temp[k] = a[i];
                k --;
                i --;
            }
        }
        while(i >= left) {
            
            temp[k] = a[i];
            i --;
            k --;
        }
        while( j > mid) {
            
            temp[k] = a[j];
            j --;
            k --;
        }
        k = 0;
        for(int n = left; n <= right; n ++)
            a[n] = temp[k ++];
    }
    public static void main(String[] args) {
        
        Solution obj = new Solution();
        int[] a = {};
        obj.InversePairs(a);
        for(int i : a)
            System.out.println(i);
        System.out.println(obj.InversePairs(a));
    }
}