leetcode315 Count of Smaller Numbers After Self

思路：

bit + 离散化。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 class Solution
 {
 public:
     int sum(vector<int> & bit, int i)
     {
         int ans = ;
         while (i) { ans += bit[i]; i -= i & -i; }
         return ans;
     }
     void add(vector<int> & bit, int i, int x)
     {
         int n = bit.size() - ;
         while (i <= n) { bit[i] += x; i += i & -i; }
     }
     vector<int> countSmaller(vector<int>& nums)
     {
         int n = nums.size();
         vector<int> a(nums.begin(), nums.end()), b(nums.begin(), nums.end());
         sort(a.begin(), a.end());
         a.erase(unique(a.begin(), a.end()), a.end());
         for (int i = ; i < n; i++)
             b[i] = lower_bound(a.begin(), a.end(), b[i]) - a.begin() + ;
         vector<int> bit(n + , ), ans(n, );
         for (int i = n - ; i >= ; i--)
         {
             add(bit, b[i], );
             ans[i] = b[i] ==  ?  : sum(bit, b[i] - );
         }
         return ans;
     }
 };
 int main()
 {
     vector<int> v;
     v.push_back();
     v.push_back(-);
     v.push_back();
     v.push_back();
     vector<int> ans = Solution().countSmaller(v);
     for (auto it: ans) cout << it << " ";
     cout << endl;
     return ;
 }