303. Range Sum Query - Immutable

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

构造函数中求出前i个数的和；

class NumArray {
public:
    NumArray(vector<int> &nums) {
        arr.push_back(0);
        for(auto num:nums){
            arr.push_back(arr.back() + num);
        }
    }

    int sumRange(int i, int j) {
        return arr[j + 1] - arr[i];
    }
private:
    vector<int> arr;
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);

class NumArray(object):
    def __init__(self, nums):
        """
        initialize your data structure here.
        :type nums: List[int]
        """
        self.accu = [0]
        for num in nums:
            self.accu.append(self.accu[-1] + num)

    def sumRange(self, i, j):
        """
        sum of elements nums[i..j], inclusive.
        :type i: int
        :type j: int
        :rtype: int
        """
        return self.accu[j + 1] - self.accu[i]


# Your NumArray object will be instantiated and called as such:
# numArray = NumArray(nums)
# numArray.sumRange(0, 1)
# numArray.sumRange(1, 2)