315. 计算右侧小于当前元素的个数

给定一个整数数组 nums,按要求返回一个新数组 counts。数组 counts 有该性质: counts[i] 的值是  nums[i] 右侧小于 nums[i] 的元素的数量。

示例:

输入: [5,2,6,1]

输出: [2,1,1,0]

解释:

5 的右侧有 2 个更小的元素 (2 和 1).

2 的右侧仅有 1 个更小的元素 (1).

6 的右侧有 1 个更小的元素 (1).

1 的右侧有 0 个更小的元素.

//分治法
class Solution {

public:

    vector countSmaller(vector& nums) {

        vector> vec;

        vector count;

        for(int i = 0;i < nums.size();i++)

        {

            vec.push_back(make_pair(nums[i],i));

            count.push_back(0);

        }

        merget_sort(vec,count);

        return count;

    }

    void merget_sort(vector>& vec,vector& count)

    {

        if(vec.size() < 2)

            return;

        int mid = vec.size()/2;

        vector> sub_vec1;

        vector> sub_vec2;

        for(int i = 0;i < mid;i++)

            sub_vec1.push_back(vec[i]);

        for(int i = mid;i < vec.size();i++)

            sub_vec2.push_back(vec[i]);

        merget_sort(sub_vec1,count);

        merget_sort(sub_vec2,count);

        vec.clear();

        merget_sort_two_vec(sub_vec1,sub_vec2,vec,count);

    }

    void merget_sort_two_vec(vector>& sub_vec1,vector>& sub_vec2,vector>& vec,vector& count)

    {

        int i = 0;

        int j = 0;

        while(i < sub_vec1.size() && j < sub_vec2.size())

        {

            if(sub_vec1[i].first <= sub_vec2[j].first)

            {

                count[sub_vec1[i].second] += j;

                vec.push_back(sub_vec1[i]);

                i++;

            }

            else

            {

                vec.push_back(sub_vec2[j]);

                j++;

            }

        }

        for(;i < sub_vec1.size();i++)

        {

            count[sub_vec1[i].second] += j;

            vec.push_back(sub_vec1[i]);

        }

        for(;j < sub_vec2.size();j++)

        {

            vec.push_back(sub_vec2[j]);

        }

    }

};


//二叉搜索树

class Solution {

public:

    struct BSTNode{

        int val;

        int count;

        BSTNode* left;

        BSTNode* right;

        BSTNode(int x):val(x),left(NULL),right(NULL),count(0){}

    };

    vector countSmaller(vector& nums) {

        vector vec;

        vector count;

        vector node_vec;

        for(int i = nums.size() - 1;i >= 0;i--)

        {

            node_vec.push_back(new BSTNode(nums[i]));

        }

        count.push_back(0);

        for(int i = 1;i < node_vec.size();i++)

        {

            int count_small = 0;

            insert(node_vec[0],node_vec[i],count_small);

            count.push_back(count_small);

        }

        for(int i = node_vec.size() - 1;i >= 0;i--)

        {

            delete node_vec[i];

            vec.push_back(count[i]);

        }

        return vec;

    }

    void insert(BSTNode* node,BSTNode* insert_node,int &count_small)

    {

        if(insert_node->val <= node->val)

        {

            node->count++;

            if(node->left)

                insert(node->left,insert_node,count_small);

            else

                node->left = insert_node;

        }

        else

        {

            count_small += node->count + 1;

            if(node->right)

                insert(node->right,insert_node,count_small);

            else

                node->right = insert_node;

        }

    }

};

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