# 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.

## Analysis

参考：[https://leetcode.com/problems/range-sum-query-immutable/solution/](https://leetcode.com/problems/range-sum-query-immutable/solution/)

重点在于`sumRange()` 需要较高的效率，比如O(1)的时间复杂度。因此在初始化时要进行处理，最直觉的就是关于range sum，可以新建数组存入`[0, i)`的`sums`，这样需要`[i, j]`的时候，`sums[j + 1] - sums[i]`即可。小技巧是生成nums.length + 1的sums\[]，而`sums[i]`代表了`[0, i)`，也就是不包含`i`的和。

## Solution

```java
class NumArray {
    private int[] sums;
    public NumArray(int[] nums) {
        if (nums.length == 0) return;
        sums = new int[nums.length];
        sums[0] = nums[0];
        for (int i = 1; i < nums.length; i++) {
            sums[i] = sums[i - 1] + nums[i];
        }
    }

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