Input: [5,2,6,1]

Output: [2,1,1,0]

Explanation:

To the right of 5 there are 2 smaller elements (2 and 1).

To the right of 2 there is only 1 smaller element (1).

To the right of 6 there is 1 smaller element (1).

To the right of 1 there is 0 smaller element.

// Binary Search

class Solution {

public:

    vector<int> countSmaller(vector<int>& nums) {

        vector<int> t, res(nums.size());

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

            int left = , right = t.size();

            while (left < right) {

                int mid = left + (right - left) / ;

                if (t[mid] >= nums[i]) right = mid;

                else left = mid + ;

            }

            res[i] = right;

            t.insert(t.begin() + right, nums[i]);

        }

        return res;

    }

};

// Insert Sort

class Solution {

public:

    vector<int> countSmaller(vector<int>& nums) {

        vector<int> t, res(nums.size());

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

            int d = distance(t.begin(), lower_bound(t.begin(), t.end(), nums[i]));

            res[i] = d;

            t.insert(t.begin() + d, nums[i]);

        }

        return res;

    }

};

再来看一种利用二分搜索树来解的方法，构造一棵二分搜索树，稍有不同的地方是需要加一个变量 smaller 来记录比当前结点值小的所有结点的个数，每插入一个结点，会判断其和根结点的大小，如果新的结点值小于根结点值，则其会插入到左子树中，此时要增加根结点的 smaller，并继续递归调用左子结点的 insert。如果结点值大于根结点值，则需要递归调用右子结点的 insert 并加上根结点的 smaller，并加1，参见代码如下：

解法三：

// Binary Search Tree

class Solution {

public:

    struct Node {

        int val, smaller;

        Node *left, *right;

        Node(int v, int s) : val(v), smaller(s), left(NULL), right(NULL) {}

    };

    int insert(Node*& root, int val) {

        if (!root) return (root = new Node(val, )), ;

        if (root->val > val) return root->smaller++, insert(root->left, val);

        return insert(root->right, val) + root->smaller + (root->val < val ?  : );

    }

    vector<int> countSmaller(vector<int>& nums) {

        vector<int> res(nums.size());

        Node *root = NULL;

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

            res[i] = insert(root, nums[i]);

        }

        return res;

    }

};