303. Range Sum Query - Immutable

Easy
Given an integer array nums, handle multiple queries of the following type:
  1. 1.
    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]).
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:
  • 1 <= nums.length <= 10^4
  • -10^5 <= nums[i] <= 10^5
  • 0 <= left <= right < nums.length
  • At most 10^4 calls will be made to sumRange.

解題

type NumArray struct {
arr []int
}
​
​
func Constructor(nums []int) NumArray {
prefixSum := make([]int, len(nums))
prefixSum[0] = nums[0]
for i:=1; i<len(nums); i++ {
prefixSum[i] = prefixSum[i - 1] + nums[i]
}
return NumArray{ arr: prefixSum }
}
​
​
func (this *NumArray) SumRange(left int, right int) int {
if left != 0 {
return this.arr[right] - this.arr[left - 1]
}
return this.arr[right]
}
​
​
/**
* Your NumArray object will be instantiated and called as such:
* obj := Constructor(nums);
* param_1 := obj.SumRange(left,right);
*/