leetcode 315. Count of Smaller Numbers After Self 这个AC的做法实在是难以想到

You 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 element (1).
To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0].

这道题题意很简单，最直接的方法就是暴力求解，但是肯定会超时。

想了好久，没有想到方法，网上看到了一个方法，我思考了许久，觉得一般人想不到这个方法，包括我，所以还是暴力求解吧！

代码如下：

import java.util.ArrayList;import java.util.Arrays;import java.util.List;class Node {    Node left, right;    int val, sum, dup = 1;    public Node(int v)     {        val = v;    }}/* * https://segmentfault.com/a/1190000008233819 * 多多考虑吧，我觉得一般想不到 * */public class Solution {    public List<Integer> countSmaller(int[] nums)    {        Integer[] ans = new Integer[nums.length];        Node root = null;        for (int i = nums.length - 1; i >= 0; i--)             root = insert(nums[i], root, ans, i, 0);        return Arrays.asList(ans);    }    private Node insert(int num, Node node, Integer[] ans, int i, int preSum)     {        if (node == null)         {            node = new Node(num);            ans[i] = preSum;        } else if (node.val == num)         {            node.dup++;            ans[i] = preSum + node.sum;        } else if (node.val > num)         {            node.sum++;            node.left = insert(num, node.left, ans, i, preSum);        } else             node.right = insert(num, node.right, ans, i, preSum + node.dup + node.sum);        return node;    }    /*     * 暴力求解，但是会超时     * */    public List<Integer> countSmallerByBruceFore(int[] nums)     {        List<Integer> res=new ArrayList<>();        if(nums==null ||nums.length<=0)            return res;        for(int i=0;i<nums.length;i++)        {            int count=0;            for(int j=i+1;j<nums.length;j++)            {                if(nums[j]<nums[i])                    count++;            }            res.add(count);        }               return res;    }}

下面是C++的做法，我想来想去感觉暴力做法最简单明了，

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>using namespace std;class Solution{public:    vector<int> countSmaller(vector<int>& nums)    {        vector<int> res(nums.size(), 0);        for (int i = 0; i < nums.size(); i++)        {            for (int j = i + 1; j < nums.size(); j++)            {                if (nums[j] < nums[i])                    res[i]++;            }        }        return res;    }};