LeetCode #303: Range Sum Query - Immutable (C/C++).

easy

source: https://leetcode.com/problems/range-sum-query-immutable/
C/C++ Solution to LeetCode problem 303. Range Sum Query - Immutable.

Problem

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Given an integer array nums, handle multiple queries of the following type:

Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

• NumArray(int[] nums) Initializes the object with the integer array nums.
• int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).

Examples

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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 (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraints

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• 1 <= nums.length <= 104
• -105 <= nums[i] <= 105
• 0 <= left <= right < nums.length
• At most 104 calls will be made to sumRange.

Solution

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class NumArray {
private:
  vector<int> *n;
public:
  NumArray(vector<int>& nums) {
    n = &nums;
  }
    
  int sumRange(int left, int right) {
    int sum = 0;
    for(int i=left; i<=right; i+=1)
      sum += n->at(i);

    return sum;
  }
};