Given an integer arraynums, find the sum of the elements between indicesiandj(i≤j), inclusive.

Example:
Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

题目非常简单，所以最暴力的方法就是从数组i位置开始累加到j位置，但是这样会超时。

优化后的方法是：遍历nums,每个元素的位置存储的值是此元素之前的所有元素值之和。当计算i与j之间的元素和时，只需要用j位置的元素和减去i-1位置的元素和，并且注意考虑边界条件

public class NumArray {
    private int[] nums;
    public NumArray(int[] nums) {
        this.nums = nums;
        for(int i = 1; i < this.nums.length; i++){
            this.nums[i] += this.nums[i-1];
        }
    }
    public int sumRange(int i, int j) {
        if(i > j)
            return -1;
        else if(i == 0)
            return this.nums[j];
        else
            return this.nums[j] - this.nums[i-1];
    }
}```

/**

• Your NumArray object will be instantiated and called as such:
• NumArray obj = new NumArray(nums);
• int param_1 = obj.sumRange(i,j);
  */```