数组中的逆序对

public int reversePairs(int[] nums) {
        int num=0;
        for(int i=1;i<nums.length;i++){
            for(int j=0;j<i;j++){
                if(nums[i]<nums[j]){
                    num++;
                }
            }
        }
        return num;
    }

 public int  reversePairs(int[] nums) {

        int len=nums.length
        if (len < 2) {
            return 0;
        }

        int[] tem = new int[nums.length];
        return merge(nums,0,  len- 1, tem);
    }

    public int merge(int[] nums, int left, int right, int[] tem) {
        if (left == right) {
            return 0;
        }

        //int mid=(left+right)/2;当数字大的时候会出现整型溢出
        int mid = left + (right - left) / 2;

        int leftpart = merge(nums, left, mid, tem);
        int rightpart = merge(nums, mid + 1, right, tem);

        int part = mergeSort(nums, left, mid, right, tem);
        return leftpart + rightpart + part;

    }

    public int mergeSort(int[] nums, int left, int mid, int right, int[] tem) {
        for (int i = left; i <= right; i++) {
            tem[i] = nums[i];
        }

        int i = left;
        int j = mid + 1;

        int count = 0;//记录逆序对数量

        for (int k = left; k <= right; k++) {

            if (i == mid + 1) { //左侧到达边界
                nums[k] = tem[j];
                j++;
            } else if (j == right + 1) { // 右侧到达边界
                nums[k] = tem[i];
                i++;
            } else if (tem[i] <= tem[j]) { //当左侧的数小于右侧的数时，将该元素放入答案中
                nums[k] = tem[i];
                i++;
            } else {
                nums[k] = tem[j];
                j++;
                count += (mid + 1 - i);
            }
        }
        return count;

    }