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;