[LeetCode]Count of Range Sum

Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.
Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive.

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

Example:
Given nums = [-2, 5, -1], lower = -2, upper = 2,
Return 3.

The three ranges are : [0, 0], [2, 2], [0, 2] and their respective sums are: -2, -1, 2.

首先求出把数i的前面所有数的累加值数组，prefix_sums[i] , 那么S(i,j) = prefix_sums[j] - prefix_sums[i].

这样问题就可以简化为.

对于一个i,(0<i<n-1)，找到有效的点(j>i)使得：

lower + prefix_sums[i] <= prefix_sums[j] <= upper + prefix_sums[i]

这样可以转化为一个Range Search 问题。

这个问题可以在O(log n + K)(K是在Range个数）利用二叉搜索树解决。

STL的multiset是利用二叉搜索树实现的。利用multiset 即可。

class Solution {public:    int countRangeSum(vector<int>& nums, int lower, int upper) {        multiset<long long> Set ;        vector<long long> PreSum(nums.size()+1) ;        long long sum = 0;        PreSum[0] = 0;        for(int i=0; i<nums.size(); ++i){            sum += nums[i];            PreSum[i+1] = sum;        }        int count = 0;        for(int i=nums.size(); i>=1 ; --i){            Set.insert(PreSum[i]);                        auto lowbound = Set.lower_bound(PreSum[i-1]+lower);            auto uppbound = Set.upper_bound(PreSum[i-1]+upper);                        count += distance(lowbound,uppbound);        }        return count;    }};