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.

题意：
求数组区间和，并能够修改数组元素

思路：
直接模拟时间复杂度为$O(n^2)$，必然不能通过。题目要求数组区间和并要修改元素值，这正好是树状数组的两个特性，可以用树状数组求解。时间复杂度为$O(nlog(n))$。具体思路见树状数组。

class NumArray {
public:
    NumArray(vector<int> &nums) { //建立树状数组,注意树状数组的下标要从1开始。bit[0]中不保存元素。
        num.resize(nums.size() + 1);
        bit.resize(nums.size() + 1);
        for (int i = 0; i < nums.size(); ++i) {
            update(i, nums[i]);
        }
    }
    void update(int i, int val) {//修改元素值的函数
        int diff = val - num[i + 1];
        for (int j = i + 1; j < num.size(); j += (j&-j)) {
            bit[j] += diff;
        }
        num[i + 1] = val;
    }

    int sumRange(int i, int j) {//求区间和的函数
        return getSum(j + 1) - getSum(i);
    }
    int getSum(int i) {//得到下标从0至i-1元素的区间和
        int res = 0;
        for (int j = i; j > 0; j -= (j&-j)) {
            res += bit[j];
        }
        return res;
    }

private:
    vector<int> num;
    vector<int> bit;
};

知识点：
树状数组