给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。

update(i, val) 函数可以通过将下标为 i 的数值更新为 val，从而对数列进行修改。

示例:

Given nums = [1, 3, 5]

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

说明:

1. 数组仅可以在 update 函数下进行修改。
2. 你可以假设 update 函数与 sumRange 函数的调用次数是均匀分布的。

使用线段树的方法解决，实际复杂读是O(logN)

class NumArray {
private:
    vector<int> arr;
    vector<int> tree;
public:
    void build_tree(int node, int start, int end) {
	if (start == end) {
		tree[node] = arr[start];
		return;
	}

	int mid = start + (end - start) / 2;
	int left_node = node * 2 + 1;
	int right_node = node * 2 + 2;
	build_tree(left_node, start, mid);
	build_tree(right_node, mid + 1, end);
	tree[node] = tree[left_node] + tree[right_node];
}
    
    void update_tree(int node, int start, int end, int idx, int val) {

	if (start == end) {
		arr[idx] = val;
		tree[node] = val;
		return;
	}
	int mid = start + (end - start) / 2;
	int left_node = node * 2 + 1;
	int right_node = node * 2 + 2;
	if (idx >= start && idx <= mid)
		update_tree(left_node, start, mid, idx, val);
	else
		update_tree(right_node, mid + 1, end, idx, val);
	tree[node] = tree[left_node] + tree[right_node];
}
    
    int query_tree(int node, int start, int end, int L, int R) {
	if (R<start || L > end)
		return 0;
	else if (L <= start && end <= R)
		return tree[node];
	else if (start == end)
		return tree[node];
	int mid = start + (end - start) / 2;
	int left_node = node * 2 + 1;
	int right_node = node * 2 + 2;
	int sum_left = query_tree(left_node, start, mid, L, R);
	int sum_right = query_tree(right_node, mid+1, end, L, R);
	return sum_left + sum_right;
}

    
    
    NumArray(vector<int> nums) {
        if(nums.size()==0)
            return;
        arr.assign(nums.begin(), nums.end());
        tree.resize(nums.size()*4);
        build_tree(0,0,nums.size()-1);
        
    }
    
    void update(int i, int val) {
        update_tree(0, 0, arr.size()-1, i, val);
    }
    
    int sumRange(int i, int j) {
        return query_tree(0,0,arr.size()-1,i,j);
    }
};

/**
 * 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);
 */