Range Sum Query

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:
Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
Note:
You may assume that the array does not change.
There are many calls to sumRange function.

自己首先写了一个方法：

class NumArray {public:    NumArray(vector<int> nums) {        int res = 0;        for(int i = 0 ; i < nums.size(); ++i){            p_num.push_back(nums[i]);            res+=nums[i];            s_num.push_back(res);        }    }    int sumRange(int i, int j) {        return s_num[j] - s_num[i] + p_num[i];    }private:    vector<int> p_num;    vector<int> s_num;};

然而这种方法效率低，开辟了多个空间，又要对其赋值。 较好的方法如下:

class NumArray {public:    NumArray(vector<int> &nums) : psum(nums.size()+1, 0) {        partial_sum( nums.begin(), nums.end(), psum.begin()+1);    }    int sumRange(int i, int j) {        return psum[j+1] - psum[i];    }private:    vector<int> psum;    };