# [315. Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/)

# 思路
给定一个数组,计算每个数字右边所有小于这个数字的个数。

## 思路一、BST

我们从后往前遍历数组,如果某个数字的右边所有数字都是有序的,那么我们就可以使用二分计算该数字右边所有小于这个数字的个数,但是如何维护有序呢,如果使用插入排序,那每次维护有序数组的时间复杂度为O(n),所以总的复杂度为O(n^2),这和暴力法是一样的。

除了有序数组之外,二叉搜索树也可以进行二分。而每次向二叉树里插入元素的复杂度平均为O(logn),所以总的时间复杂度平均就为O(nlogn)。另外,每个节点需要存放以这个节点为根的树有多少个节点。

时间复杂度平均为O(nlogn),空间复杂度为O(n)。注意BST可能会退化成链表,这样时间复杂度就为O(n^2)了。

## 思路二、归并排序

如果某个元素`nums[i]`大于其右边的某个元素`nums[j]`(j > i),那么这元素`<i, j>`就构成了一个逆序对,所以我们只需要求出以`nums[i]`为第一个元素的逆序对个数。

求逆序对最经典的方法就是分治,即归并排序。所以这题我们也可以用归并排序,只需要新增一行代码:在进行`merge`时,记录有多少个以`nums[i]`开头的逆序对。

时间复杂度为O(nlogn),空间复杂度为O(n)。

## 思路三、线段树/树状数组

此题还可以用线段树/树状数组做,我们知道线段树和树状数组可以求前缀和,而这题可以转换成求前缀和。具体转换过程如下:

1. 我们先遍历一遍数组,确定数组中元素的最小值`MIN`和`MAX`,然后想象有一个大小为`MAX - MIN`的全0数组`arr`,在这个数组上构建线段树/树状数组;
2. 然后从后往前遍历数组nums,将`arr[nums[i]]++`,更新线段树/树状数组,这样`arr`在区间`(nums[i], MAX]`的元素和即为`nums[i]`右侧它小的元素个数。

时间复杂度O(nlogN),空间复杂度O(N),其中`N = MAX - MIN`;

# C++
## 思路一
``` C++
struct BstNode{
    int val, node_num; // node_num记录这棵树有多少节点
    BstNode *left, *right;
    BstNode(int x): val(x), node_num(1), left(NULL), right(NULL){}
};

class Solution {
private:
    void BST_insert(BstNode *root, BstNode *node){
        root -> node_num += 1;
        if(node -> val >= root -> val){ // 插入到右子树
            if(root -> right) BST_insert(root -> right, node);
            else root -> right = node;
        }
        else{ // 插入到左子树
            if(root -> left) BST_insert(root -> left, node);
            else root -> left = node;
        }
    }
    
    int count(BstNode *root, int target){
        if(!root) return 0;
        
        if(root -> val < target) 
            return 1 + (root -> left == NULL ? 0 : root -> left -> node_num) \
            + count(root -> right, target);

        else return count(root -> left, target);
    }
   
public:
    vector<int> countSmaller(vector<int>& nums) {
        vector<int>res(nums.size(), 0);
        if(nums.empty()) return res;
        BstNode *root = new BstNode(nums.back());
        for(int i = nums.size() - 2; i >= 0; i--){
            res[i] = count(root, nums[i]);
            BstNode *node = new BstNode(nums[i]);
            BST_insert(root, node);
        }
        return res;
    }
};
```

## 思路二
``` C++
class Solution {
private:
    vector<int>res;
    void merge_sort(vector<pair<int, int>>&nums_with_idx, int l, int r){
        if(l >= r) return;
        int mid = (l + r) / 2;
        merge_sort(nums_with_idx, l, mid);
        merge_sort(nums_with_idx, mid+1, r);
        merge(nums_with_idx, l, mid, r);
    }
    
    void merge(vector<pair<int, int>>&nums_with_idx, int l, int mid, int r){
        vector<pair<int, int>>merged;
        int i = l, j = mid + 1;
        while(i <= mid && j <= r){
            if(nums_with_idx[i].first <= nums_with_idx[j].first){ 
                // 与普通归并排序相比新增的一步, 即记录逆序数:
                res[nums_with_idx[i].second] += j - mid - 1; // nums[i]大于nums[mid+1,...,j-1]
                merged.push_back(nums_with_idx[i++]);
            }
            else merged.push_back(nums_with_idx[j++]);
        }
        while(i <= mid){
            res[nums_with_idx[i].second] += j - mid - 1;
            merged.push_back(nums_with_idx[i++]);
        }
        //while(j <= r) merged.push_back(nums_with_idx[j++]);
        for(int k = 0; k < merged.size(); k++) nums_with_idx[k+l] = merged[k];
    }
    
public:
    vector<int> countSmaller(vector<int>& nums) {
        vector<pair<int, int>>nums_with_idx;
        res = vector<int>(nums.size(), 0);
        for(int i = 0; i < nums.size(); i++) 
            nums_with_idx.push_back({nums[i], i});
        merge_sort(nums_with_idx, 0, nums.size() - 1);
        return res;
    }
};
```

## 思路三、树状数组
[来源](https://leetcode-cn.com/problems/count-of-smaller-numbers-after-self/solution/c-shu-zhuang-shu-zu-by-mryx/)
``` C++
class Solution {
public:
    int* tree, n;
    int lowbit(int x){
        return x&(-x);
    }
    void update(int pos, int delta){
        while (pos <= n){
            tree[pos] += delta;
            pos += lowbit(pos);
        }
    }
    int getSum(int pos){
        int ret = 0;
        while (pos){
            ret += tree[pos];
            pos -= lowbit(pos);
        }
        return ret;
    }
    vector<int> countSmaller(vector<int>& nums) {
        n = nums.size();
        vector<int> ret(n);
        if (n == 0) return ret;
        
        int minn = -50000, maxx = 50000;
        for (int i=0;i<n;++i){
            maxx = max(maxx, nums[i]);
            minn = min(minn, nums[i]);
        }
        n = maxx - minn + 2;
        tree = new int[n+1];
        memset(tree, 0, sizeof(int)*n);
        for (int i=nums.size()-1;i>=0;--i){
            ret[i] = getSum(nums[i] - minn);
            update(nums[i]-minn+1, 1);
        }
        return ret;
    }
};
```

## 思路三、线段树
[来源](https://leetcode-cn.com/problems/count-of-smaller-numbers-after-self/solution/c-xian-duan-shu-jie-fa-by-dufre/)

``` C++
struct SegmentTreeNode{
        int start;
        int end;
        int count;

        SegmentTreeNode* left;
        SegmentTreeNode* right;

        SegmentTreeNode(int _start, int _end):start(_start),end(_end) {
            count = 0;
            left = NULL;
            right = NULL;
        }
};

class Solution {
public:
    SegmentTreeNode* build(int start, int end){
        if (start > end)
            return NULL;
        
        SegmentTreeNode* root = new SegmentTreeNode(start, end);

        if (start == end){
            root->count = 0;
        }else{
            int mid = start + (end - start)/2;
            root->left = build(start, mid);
            root->right = build(mid+1, end);
        }
        return root;
    }

    int count(SegmentTreeNode* root, int start, int end){
        if (root == NULL || start>end)
            return 0;
        if (start==root->start && end==root->end){
            return root->count;
        }
        int mid = root->start + (root->end - root->start)/2;
        int leftcount = 0, rightcount = 0;

        if (start <= mid){
            if (mid < end)
                leftcount = count(root->left, start, mid);
            else
                leftcount = count(root->left, start, end);
        }

        if (mid < end){
            if (start <= mid)
                rightcount = count(root->right, mid+1, end);
            else
                rightcount = count(root->right, start, end);
        }

        return (leftcount + rightcount);
    }

    void insert(SegmentTreeNode* root, int index, int val){
        if (root->start==index && root->end==index){
            root->count += val;
            return;
        }

        int mid = root->start + (root->end - root->start)/2;
        if (index>=root->start && index<=mid){
            insert(root->left, index, val);
        }
        if (index>mid && index<=root->end){
            insert(root->right, index, val);
        }

        root->count = root->left->count + root->right->count;
    }

    vector<int> countSmaller(vector<int>& nums) {
        vector<int> res;
        if (nums.empty())
            return res;
        res.resize(nums.size());
        int start = nums[0];
        int end = nums[0];

        for (int i=1; i<nums.size(); i++){
            start = min(start, nums[i]);
            end = max(end, nums[i]);
        }

        SegmentTreeNode* root = build(start, end);

        for (int i=nums.size()-1; i>=0; i--){
            res[i] = count(root, start, nums[i]-1);
            insert(root, nums[i], 1);
        }  

        return res;      
    }
};
```