303. 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:

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

Solution:

Use an array to keep track the summation until current element. We can use sum[j]-sum[i] when calculate range[i,j]. 

I use one extra size to avoid special care for range starting from index 0. 
Time complexity:
Initiate sum array: O(n)
calculate range sum: O(1).

 class NumArray {  
   int[] sum;  
   public NumArray(int[] nums) {  
     sum=new int[nums.length+1];  
     for(int i=0;i<nums.length;i++){  
       sum[i+1]=nums[i]+sum[i];  
     }  
   }  
   public int sumRange(int i, int j) {  
     return sum[j+1]-sum[i];  
   }  
 }