【剑指Offer】51. 数组中的逆序对(归并排序/树状数组)

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。

示例 1:

输入: [7,5,6,4]
输出: 5

限制：
0 <= 数组长度 <= 50000

归并排序

class Solution {
    public int reversePairs(int[] nums) {
        int len = nums.length;
        int[] copy = new int[len];
        int[] temp = new int[len];
        if(len < 2) return 0;
        for(int i = 0; i < len; i++){
            copy[i] = nums[i];
        }
        return mergeSort(copy, 0, len - 1, temp);
    }
    public int mergeSort(int[] nums, int left, int right, int[] temp){
        if(left == right) return 0;
        int mid = left + (right - left) / 2;
        int leftPairs = mergeSort(nums, left, mid, temp);
        int rightPairs = mergeSort(nums, mid + 1, right, temp);
        if(nums[mid] <= nums[mid + 1]){
            return leftPairs + rightPairs;
        }
        int crossPairs = mergeCount(nums, left, mid, right, temp);
        return leftPairs + rightPairs + crossPairs;
    }
    public int mergeCount(int[] nums, int left, int mid, int right, int[] temp){
        for(int i = left; i <= right; i++){
            temp[i] = nums[i];
        }
        int i = left;
        int j = mid + 1;
        int count = 0;
        for(int k = left; k <= right; k++){
            if(i == mid + 1){
                nums[k] = temp[j];
                j++;
            }else if(j == right + 1){
                nums[k] = temp[i];
                i++;
            }else if(temp[i] <= temp[j]){
                nums[k] = temp[i];
                i++;
            }else{
                nums[k] = temp[j];
                j++;
                count += (mid - i + 1);
            }
        }
        return count;
    }
}

树状数组

待补