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.

Solution："积分" 累积求和

思路：
Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

class NumArray {

    int[] nums;

    public NumArray(int[] nums) {
        for(int i = 1; i < nums.length; i++)
            nums[i] += nums[i - 1];

        this.nums = nums;
    }

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

        return nums[j] - nums[i - 1];
    }
}