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:

You may assume that the array does not change.

There are many calls to sumRange function.

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

   arr[i]是nums数组从0到i（包括）的sum，所以要求i，j之间的和，先判断i是否为0，然后等于arr[j] - arr[i - 1]。

以上是dp解法。