303. Range Sum Query – Immutable-区域和检索 – 数组不可变

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

Implement the NumArray class:

• NumArray(int[] nums) Initializes the object with the integer array nums.
• int sumRange(int i, int j) Return the sum of the elements of the nums array in the range [i, j] inclusive (i.e., sum(nums[i], nums[i + 1], … , nums[j]))

Example 1:
Input
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
Output
[null, 1, -1, -3]

Explanation
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1))
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

Constraints:

• 0 <= nums.length <= 10^4
• -10^5 <= nums[i] <= 10^5
• 0 <= i <= j < nums.length
• At most 104 calls will be made to sumRange.

Solution in python:

class NumArray:

    def __init__(self, nums: List[int]):
        self.array = nums

    def sumRange(self, i: int, j: int) -> int:
        return sum(self.array[i:j+1])

# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)