493. Reverse Pairs

// see more at https://www.youtube.com/watch?v=j68OXAMlTM4
// https://leetcode.com/problems/reverse-pairs/discuss/97268/general-principles-behind-problems-similar-to-reverse-pairs
// http://www.cnblogs.com/grandyang/p/6657956.htmlclass Solution {
public:
    int reversePairs(vector<int>& nums) {
        return reversePairs(nums, , nums.size()-);
    }
    int reversePairs(vector<int>& nums, int begin, int end) {
        if (begin >= end)   return ;
        int m = begin + (end - begin) / ;
        int res = reversePairs(nums, begin, m);
        res += reversePairs(nums, m+, end);        int *p = new int[end-begin+];        int i = begin, j = m+;
        while (i <= m && j <= end) {
            if (nums[i] > 2L * nums[j]) {   // avoid overflow
                res += m - i + ;
                j++;
            }
            else {
                i++;
            }
        }        i = begin; j = m+; int k = ;
        while (i <= m && j <= end) {
            if (nums[i] > nums[j])
                p[k++] = nums[j++];
            else
                p[k++] = nums[i++];
        }
        while (i <= m)
            p[k++] = nums[i++];
        while (j <= end)
            p[k++] = nums[j++];
        for (int i = begin; i <= end; i++) {
            nums[i] = p[i-begin];
        }
        //sort(nums.begin() + begin, nums.begin() + end+1);        return res;
    }
};