leetcode 315. Count of Smaller Numbers After Self

/*leetcode 315. Count of Smaller Numbers After SelfYou are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].Example:    Given nums = [5, 2, 6, 1]    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 To (1).    element the right of 1 there is 0 smaller element.Return the array [2, 1, 1, 0].题目大意：给定一个数组，返回一个相同大小的数组，数组中每个位置的元素为原数组中后面小于该元素的个数。解题思路：(1)遍历，比较(TLE)(2)从右向左遍历数组，同时维护一个数组，每次从这个数组中找到第一个大于当前元素的位置求得右边有多少个比它小的数。时间复杂度为O(n^2)(3)对(2)的改进：查找时用二分的方法*/#include <iostream>#include <vector>#include <algorithm>using namespace std;class Solution {public:    //TLE    vector<int> countSmaller(vector<int>& nums)     {        int len = nums.size();        if (len == 0)            return nums;        vector<int> ret(len, 0);        for (int i = 0; i < len; ++i)        {            for (int j = i + 1; j < len; ++j)            {                if (nums[i] > nums[j])                    ++ret[i];            }        }        return ret;    }    //维护一个数组，是遍历过的nums中的一个有序的数组    vector<int> countSmaller1(vector<int>& nums)    {        int len = nums.size();        if (len == 0)            return nums;        vector<int> v;//维护的数组        vector<int> ret(len, 0);        for (int i  = len - 1; i >= 0; --i)        {            auto iter = lower_bound(v.begin(), v.end(), nums[i]);            ret[i] = iter - v.begin();            v.insert(iter, nums[i]);        }        return ret;    }    vector<int> countSmaller2(vector<int>& nums)    {        int len = nums.size();        if (len == 0)            return nums;        vector<int> v;//维护的数组        vector<int> ret(len, 0);        for (int i = len - 1; i >= 0; --i)        {            if (v.empty())            {                ret[i] = 0;                v.push_back(nums[i]);            }            else            {                //二分查找:找到比nums[i]                int l = 0, r = v.size() - 1;                while (l <= r)                {                    int mid = (l + r) / 2;                    if (v[mid] < nums[i])                        l = mid + 1;                    else                        r = mid - 1;                }                v.insert(v.begin() + l, nums[i]);                ret[i] = l;            }        }        return ret;    }};//BST//二分查找，维护一个class Solution1 {public:    //node    typedef struct Node {        int val, smaller;//the current node value and the count smaller then val int the child        struct Node* left, *right;        Node(int v, int s) : val(v), smaller(s), left(NULL),right(NULL){}    }node;    int insert(node* &root, int v)    {        if (root == NULL) {            root = new node(v, 0);            return 0;        }        if (root->val > v) {            root->smaller++;            return insert(root->left, v);        }        else {            return (insert(root->right, v) + root->smaller + (root->val < v ? 1: 0));        }    }    vector<int> countSmaller(vector<int>& nums)    {        vector<int> ret(nums.size(), 0);        node* root = NULL;        for (int i = nums.size()-1;i >=0; --i) {            ret[i] = insert(root, nums[i]);            cout << ret[i] << " " << endl;        }        cout << endl;        return ret;    }};void test(){    vector<int> v{5,2,6,1};    Solution sol;    //vector<int> ret = sol.countSmaller(v);    //vector<int> ret = sol.countSmaller1(v);    vector<int> ret = sol.countSmaller2(v);    for (auto i : ret)        cout << i << " ";    cout << endl;    Solution1 sol1;    vector<int> ret1 = sol1.countSmaller(v);    for (auto i : ret1)        cout << i << " ";    cout << endl;}int main(){    test();    return 0;}