303. Range Sum Query - Immutable

Easy

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

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);
 */