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

Note:
You may assume that the array does not change.
There are many calls to sumRange function.

Analysis

Ref1
When initialize, store num[i] as the sum from index 0 to i.
When get query i to j, return num[j]-num[i]
Because there are many calls to sum range, this method has time complex for initialize O(n), and time complex for sumRange O(1), which is good!

Code

    public class NumArray {
    int[] nums;

    public NumArray(int[] nums) {
        for (int i=1; i<nums.length; i++) {nums[i]+=nums[i-1];}
        //this line is important!
        this.nums = nums;
    }

    public int sumRange(int i, int j) {
        if (i==0) {return nums[j];}
        return nums[j]-nums[i-1];
    }
}

Questions

why this line?
this.nums = nums;