leetcode 493. Reverse Pairs 逆序对数量 + 归并排序做法

Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].

You need to return the number of important reverse pairs in the given array.

Example1:

Input: [1,3,2,3,1]
Output: 2
Example2:

Input: [2,4,3,5,1]
Output: 3
Note:
The length of the given array will not exceed 50,000.
All the numbers in the input array are in the range of 32-bit integer.

题意很简单，但是AC的方法不会，我觉得循环遍历也不错，虽然肯定会超时

代码如下：

#include 
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#include 

using namespace std;

class Solution 
{
public:
    int reversePairs(vector<int>& nums) 
    {
        int res = sort_and_count(nums.begin(), nums.end());
        return res;
    }
    int sort_and_count(vector<int>::iterator begin, vector<int>::iterator end) 
    {
        if (end - begin <= 1)
            return 0;
        else
        {
            auto mid = begin + (end - begin) / 2;
            int count = sort_and_count(begin, mid) + sort_and_count(mid, end);
            for (auto i = begin, j = mid; i != mid; ++i) 
            {
                while (j != end && (*i) > 2L * (*j) )
                    ++j;
                count += j - mid;
            }
            inplace_merge(begin, mid, end);
            return count;
        }
    }


    int reversePairsByLoop(vector<int>& nums)
    {
        int count = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            for (int j = i + 1; j < nums.size(); j++)
            {
                long long a = nums[i];
                long long b = ((long long)nums[j]) << 1;
                if (a > b)
                    count++;
            }
        }
        return count;
    }
};

可以按照逆序对的做法来做，参考这个博客剑指offer读书笔记：第五章，优化时间和空间效率

#include 
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using namespace std;


class Solution
{
public:
    int reversePairs(vector<int>& a)
    {
        if (a.size() <= 0)
            return 0;
        vector<long long> all(a.begin(), a.end());
        return mergeSort(all, 0, all.size() - 1);
    }

    long long mergeSort(vector<long long>& a, int begin, int end)
    {
        if (begin == end)
            return 0;
        else
        {
            long long mid = (end - begin) / 2 + begin;
            long long leftCount = mergeSort(a, begin, mid);
            long long rightCount = mergeSort(a, mid + 1, end);
            long long midCount = 0;
            long long i = begin, j = mid + 1;
            while (i <= mid && j <= end)
            {
                if (a[i] <= 2 * a[j])
                    i++;
                else
                {
                    midCount += (mid - i + 1);
                    j++;
                }
            }

            vector<long long> tmp;
            i = begin, j = mid + 1;
            while (i <= mid && j <= end)
            {
                if (a[i] <= a[j])
                    tmp.push_back(a[i++]);
                else
                    tmp.push_back(a[j++]);
            }
            while (i <= mid)
                tmp.push_back(a[i++]);
            while (j <= end)
                tmp.push_back(a[j++]);

            for (int i = begin; i <= end; i++)
                a[i] = tmp[i - begin];
            return leftCount + midCount + rightCount;
        }
    }
};