leetcode Range Sum Query - Mutable

原题链接：https://leetcode.com/problems/range-sum-query-mutable/

Description

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.

线段树单点更新。。

class NumArray {public:    NumArray() = default;    NumArray(vector<int> &nums) {        n = (int)nums.size();        if (!n) return;        arr = new int[n << 2];        built(1, 1, n, nums);    }    ~NumArray() { delete []arr; }    void update(int i, int val) {        update(1, 1, n, i + 1, val);    }    int sumRange(int i, int j) {        return sumRange(1, 1, n, i + 1, j + 1);    }private:    int n, *arr;    void built(int root, int l, int r, vector<int> &nums) {        if (l == r) {            arr[root] = nums[l - 1];            return;        }        int mid = (l + r) >> 1;        built(root << 1, l, mid, nums);        built(root << 1 | 1, mid + 1, r, nums);        arr[root] = arr[root << 1] + arr[root << 1 | 1];    }    void update(int root, int l, int r, int pos, int val) {        if (pos > r || pos < l) return;        if (pos <= l && pos >= r) {            arr[root] = val;            return;        }        int mid = (l + r) >> 1;        update(root << 1, l, mid, pos, val);        update(root << 1 | 1, mid + 1, r, pos, val);        arr[root] = arr[root << 1] + arr[root << 1 | 1];    }    int sumRange(int root, int l, int r, int x, int y) {        if (x > r || y < l) return 0;        if (x <= l && y >= r) return arr[root];        int mid = (l + r) >> 1, ret = 0;        ret += sumRange(root << 1, l, mid, x, y);        ret += sumRange(root << 1 | 1, mid + 1, r, x, y);        return ret;    }};