NowCoder 數組中的逆序對 歸併排序

題意：給一個數組，求出數組中存在的逆序對的個數，逆序對即兩數中前面的數字大於後面的數字
思路：暴力枚舉很簡單，複雜度爲O(n²)，可以用歸併排序的性質，歸併後兩個子數組中設前面爲i，後面爲j，若array[i]>array[j]，則i到mid的數即爲當前可知逆序對數（排序完後前面數組i到mid的數都比i位置的大，故排序前亂序時前面數組相對後面數組逆序對也有這麼多），將所有子數組的逆序對統計即爲數組的總逆序對個數

public class Solution {
    public int InversePairs(int [] array) {
        return solve(array, 0, array.length-1);
    }
    public int solve(int[] array, int l, int r) {
        if (l >= r)
            return 0;
        int mid = (l+r)>>1;
        int left = solve(array, l, mid);
        int right = solve(array, mid+1, r);
        return (left+right+merge(array, l, r))%1000000007;
    }
    public int merge(int[] array, int l, int r) {
        int[] temp = new int[r-l+1];
        int i = l, mid = (l+r)>>1, j = mid + 1, cnt = 0, cur = 0;
        while (i <= mid || j <= r) {
            if (j > r)
                temp[cur++] = array[i++];
            else if (i > mid)
                temp[cur++] = array[j++];
            else if (array[i] > array[j]) {
                cnt = (cnt+mid-i+1)%1000000007;
                temp[cur++] = array[j++];
            } else {
                temp[cur++] = array[i++];
            }
        }
        for (int k = 0; k < temp.length; k++)
            array[k+l] = temp[k];
        return cnt;
    }
}