307. Range Sum Query - Mutable

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

The update(i, val) function modifies nums by updating the element at index i to val.
Example:

Given nums = [1, 3, 5]

sumRange(0, 2) 9
update(1, 2)
sumRange(0, 2) 8

Note:
The array is only modifiable by the update function.
You may assume the number of calls to update and sumRange function is distributed evenly.

class NumArray
    int[] processed;
    int[] nums;
    int length;

    public NumArray(int[] nums) {
        length = nums.length;
        processed = new int[length+1];
        this.nums = nums;

        //init processed
        for(int i = 1;i<=length;i++){
            int sum = 0;
            int count = 1;
            int counter = lowBit(i);

            while(count <= counter){
                sum += nums[i-count];
                count++;
            }
            processed[i] = sum;
        }
    }

    void update(int i, int val) {
        //更新树状数组
        int gap = val - nums[i];
        nums[i] = val;

        int index = i+1;
        while(index <= length){
            processed[index] += gap;
            index += lowBit(index);
        }
    }

    public int sumRange(int i, int j) {
        return sum(j+1) - sum(i);
    }

    private int sum(int index){
        int sum = 0;
        while(index > 0){
            sum += processed[index];
            index -= lowBit(index);
        }
        return sum;
    }

    private int lowBit(int index){
        return index & (-index);
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * obj.update(i,val);
 * int param_2 = obj.sumRange(i,j);
 */