問題：

給定一個數組，求連續元素之和在給定範圍[lower, upper]之間的，連續idx為(i～j)元素組個數。

Note:
A naive algorithm of O(n2) is trivial. You MUST do better than that.

Example:
Input: nums = [-2,5,-1], lower = -2, upper = 2,
Output: 3 
Explanation: The three ranges are : [0,0], [2,2], [0,2] and their respective sums are: -2, -1, 2.
 

Constraints:
0 <= nums.length <= 10^4

解法：FenwickTree

解法類似315. Count of Smaller Numbers After Self

• 只是需要多求一步presum->sum，作為處理物件。
• 對字首和陣列，進行sort->sortsum，來確定在FenwickTree中的idx。
• FenwickTree記錄對應相同sum值的個數（出現頻率）。

然後對原順序的sum，逐一讀入，

每個sum[i]中，在FenwickTree中，尋找sum[i]-upper ~ sum[i]-lower 範圍內共有多少個，res追加。

然後把當前sum[i]找到idx，插入FenwickTree中，累計count+1。

程式碼參考：

 1 class FenwickTree {
 2 public:
 3     FenwickTree(int n):tree(n+1, 0) {}
 4     void update(int i, int delta) {
 5         while(i<tree.size()) {
 6             tree[i]+=delta;
 7             i+=lowbit(i);
 8         }
 9     }
10     int getPresum(int i) {
11         int sum=0;
12         while
 
(i>0){
13             sum+=tree[i];
14             i-=lowbit(i);
15         }
16         return sum;
17     }
18 private:
19     vector<int> tree;
20     int lowbit(int x) {
21         return x&(-x);
22     }
23 };
24 
25 class Solution {
26 public:
27     int countRangeSum(vector<int>& nums, int lower, int upper) {
28         int cursum=0;
29         int res=0;
30         vector<long long> sum(nums.size()+1,0);
31         for(int i=0; i<nums.size(); i++){
32             sum[i+1]=sum[i]+nums[i];
33         }
34         vector<long long> sortsum(sum);
35         sort(sortsum.begin(), sortsum.end());
36         
37         FenwickTree ftree(sum.size());
38         for(int i=0; i<sum.size(); i++){
39             int idx_low = distance(sortsum.begin(), lower_bound(sortsum.begin(),sortsum.end(),sum[i]-upper));
40             int idx_upper = distance(sortsum.begin(), upper_bound(sortsum.begin(),sortsum.end(),sum[i]-lower));
41             res+=(ftree.getPresum(idx_upper)-ftree.getPresum(idx_low));
42             
43             int idx = distance(sortsum.begin(), lower_bound(sortsum.begin(),sortsum.end(),sum[i]));
44             ftree.update(idx+1, 1);
45         }
46         return res;
47     }
48 };