题解 | #旋转数组的最小数字#

public class Solution {
    public int InversePairs(int [] array) {
       //归并排序  。
        if(array.length==0){
            return 0;
        }
        return mergeSort(array,0,array.length-1)%1000000007;
    }
    public  int merge(int[] array,int left,int mid,int right){
        int start = left;
        int mm = mid+1;
        int[] tmp = new  int[right-left+1];
        int index = 0;
        int count =0;
        while(start <= mid && mm<= right){
            if(array[start] <= array[mm]){
                tmp[index++] = array[start++];
            }else{
                count += (mid-start+1);
                tmp[index++] = array[mm++];

            }
        }
        while(start<=mid){
            tmp[index++] = array[start++];
        }
        while(mm <= right){
            tmp[index++] = array[mm++];
        }
        index=0;
        for(int i= left;i<=right;i++){
            array[i] = tmp[index++];
        }
        return count % 1000000007;
    }
    public int mergeSort(int[] array,int left,int right){
        if(left>=right){
            return 0;
        }
        int mid = (left+right)>>1;
        int leftSum = mergeSort(array,left,mid);
        int rightSum = mergeSort(array,mid+1,right);
        return (leftSum + rightSum + merge(array,left,mid,right));
    }
}

为什么不能在mergeSort最后return进行模 非要在第一个里面进行% 要不然 倒数第二个用例过不了 。。。