Reverse Pairs

For an array A, if i < j, and A [i] > A [j], called (A [i], A [j]) is a reverse pair.
return total of reverse pairs in A.

 

Example

Given A = [2, 4, 1, 3, 5] , (2, 1), (4, 1), (4, 3) are reverse pairs. return 3

分析：

如果用两个for loop,非常简单。但要降低复杂度，做法非常变态.简单来说，原理就是利用merge sort的方法，把左subarray 和右subarray 合并（假设它们都已排好序）。然后比较当前左subarray的第一个值是否比右边subarray第一个值大两倍，如果是，那么很明显左subarray所有的都比那个值大两倍。

class Solution {
    public int reversePairs(int[] nums) {
        return mergeSort(nums, 0, nums.length - 1);
    }
    
    public int mergeSort(int[] nums, int start, int end) {
        if (start >= end) return 0;
        int total =  0;
        int mid = start + (end - start) / 2;
        total += mergeSort(nums, start, mid);
        total += mergeSort(nums, mid + 1, end);
        total += merge(nums, start, mid, end);
        return total;
    }

    public int merge(int[] nums, int start, int mid, int end) {
        int[] temp = new int[end - start + 1];
        int p1 = start, p2 = mid + 1;
        int index = 0,count = 0;
        
        while (p1 <= mid && p2 <= end) {
            if ((long) nums[p1] > 2 * ((long) nums[p2])) {
                count += mid - p1 + 1;
                p2++;
            } else {
                p1++;
            }
        }
        p1 = start;
        p2 = mid + 1;
        while (p1 <= mid && p2 <= end) {
            if (nums[p1] < nums[p2]) {
                temp[index++] = nums[p1++];
            } else {
                temp[index++] = nums[p2++];
            }
        }
        
        while(p1 <= mid) {
            temp[index++] = nums[p1++];
        }
        while(p2 <= end) {
            temp[index++] = nums[p2++];
        }
        System.arraycopy(temp, 0, nums, start, end - start + 1);
        return count;
    }
}