leetcode 493. Reverse Pairs

　　一道归并排序求逆序对的变体题。需要注意的是输入数据乘以2后可能超出int范围。

　　AC代码如下：

class Solution {
    vector<int> merge(vector<int> a1, vector<int> a2, int& sum)
    {
        vector<int> merged;
        int p1 = 0, p2 = 0;
        for (int i = 0,j=0; i<a1.size(); ++i)
        {
            for(;j<a2.size()&&(long long)a1[i]>2*(long long)a2[j];j++);
            sum+=j;
        }
        while (p1 != a1.size() && p2 != a2.size())
        {
            if (a1[p1]>a2[p2])
            {
                merged.push_back(a2[p2]);
                p2++;
            }
            else
            {
                merged.push_back(a1[p1]);
                p1++;
            }
        }
        while (p1 != a1.size())
        {
            merged.push_back(a1[p1]);
            p1++;
        }
        while (p2 != a2.size())
        {
            merged.push_back(a2[p2]);
            p2++;
        }
        return merged;
    }

    int mergeSort(vector<int>& nums, int l, int r)
    {
        int sum = 0;
        if (l + 1 >= r)
        {
            return 0;
        }
        else
        {
            int m = (l + r) / 2;
            sum += mergeSort(nums, l, m);
            sum += mergeSort(nums, m, r);
            vector<int> vl, vr, merged;
            for (int i = l; i<m; ++i)
            {
                vl.push_back(nums[i]);
            }
            for (int i = m; i<r; ++i)
            {
                vr.push_back(nums[i]);
            }
            merged = merge(vl, vr, sum);
            for (int i = 0; i<merged.size(); ++i)
            {
                nums[i + l] = merged[i];
            }
            return sum;
        }
    }

public:
    int reversePairs(vector<int>& nums) {
        return mergeSort(nums, 0, nums.size());
    }
};

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