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) -> 9
update(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