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

Note:

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

这题貌似简单，但你用naive的解法去解绝壁要超时。明眼人一眼就看出来了，这题要用树状数组来解决。如果你以前从没听说过什么线段树、树状数组，完全是自己想出来的，那你就是个天才。

树状数组在这里就不讲解了，自己搜去。

class NumArray {private:    vector<int> c;    vector<int> m_nums;public:    NumArray(vector<int> &nums) {        c.resize(nums.size() + 1);        m_nums = nums;        for (int i = 0; i < nums.size(); i++){            add(i + 1, nums[i]);        }    }    int lowbit(int pos){        return pos&(-pos);    }    void add(int pos, int value){        while (pos < c.size()){            c[pos] += value;            pos += lowbit(pos);        }    }    int sum(int pos){        int res = 0;        while (pos > 0){            res += c[pos];            pos -= lowbit(pos);        }        return res;    }    void update(int i, int val) {        int ori = m_nums[i];        int delta = val - ori;        m_nums[i] = val;        add(i + 1, delta);    }    int sumRange(int i, int j) {        return sum(j + 1) - sum(i);    }};

accept