Range Sum Query - Mutable

Problem

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

Solution

自己创建一个segment tree的数据结构老是超时。。

参考了官方的答案，用数组来实现一个数，挺巧妙的。

class NumArray {
    int N;
    vector<int> tree;
    int find( int left, int right) {
        if(left > right) return 0;
        if(left == right) return tree[left];
        int sum = 0;
        if( left%2 == 1 ){
            sum += tree[left++];
        } 
        if(right%2 == 0 ) {
            sum += tree[right--];
        }
        return sum += find( left/2, right/2);
    }
public:
    NumArray(vector<int> &nums): N(nums.size()) {
        
        tree.resize(2*N,0);
        for( int i = N; i < 2*N; i++){
            tree[i] = nums[i-N];
        }
        for( int i = N - 1; i > 0; i--){
            tree[i] = tree[2*i] + tree[2*i+1];
        }
    }

    void update(int idx, int val) {
        idx += N;
        int diff = val - tree[idx];
        while(idx != 0) {
            tree[idx] += diff;
            idx /= 2;
        }
    }

    int sumRange(int i, int j) {
        return find( i + N, j + N);
    }
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.update(1, 10);
// numArray.sumRange(1, 2);